![1.1] SETS AND NUMBERS 3
The laws of operation with U and fn on sets have some likeness
to those with + and x on numbers. In fact U and n have replaced
the signs for sum and product formerly used in the algebra of sets.
The likeness is only partial: the identity 4 UA = A has no analogue
in numbers; the second law in theorem 1.11 (iii) has an analogue, but
not the first.
The definitions of union and intersection previously given may be
extended. If @ is an arbitrary collection of sets, the union of these
sets, denoted by UA
Act
is the set consisting of all those elements which belong to at least one
of the sets A. The intersection of the sets of @, denoted by
MN A,
Ace
is the set consisting of the elements which belong to all the sets 4.
We require one more operation with sets. If A, B are any two sets,
the difference YER
is the set consisting of those elements which belong to A but not to B.
Note that the definition does not require B to be a subset of A. How-
ever a particularly important case occurs when all sets under con-
sideration are subsets of a given set X. Then the set X—A is called
the complement of A (relative to X) and we shall denote it by A’.
Clearly (A’)’ = A.
Theorem 1.12. For any collection @ of subsets of a set X,
COPA] fe Sands Cay) =U 42
Ac€ Ac€ Ac€ Ac€
The proof is left to the reader.
Theorem 1.12 shows that the operations U and () are comple-
mentary and that an algebraic identity involving U and (-) will admit
of a dual identity obtained by interchanging U, (). For example, in
each pair of identities in theorem 1.11 the second follows from the
first.
We shall take for granted that the reader is familiar with the systems
of real and complex numbers. There are well known and simple
methods for obtaining the complex numbers from the real numbers.
For our purposes it is only necessary to postulate the existence of the
system of real numbers as one satisfying certain axioms which are
1-2](https://image.slidesharecdn.com/22284738-250620074959-26290e84/85/A-Second-Course-In-Mathematical-Analysis-J-C-Burkill-H-Burkill-17-320.jpg)
![4 SETS AND FUNCTIONS [1.1
given in C1 (11-12) and are restated in the notes at the end of the
chapter. However, it is possible to construct the real numbers from
much more primitive objects. Again the notes contain some remarks
on the subject.
The set of real numbers will be denoted by R?, the set of complex
numbers by Z. Our notation for the open interval in R1 which consists
of the points x such that a < x < b is (a, b). The closed interval of
points x such that a < x < b is written [a, 5]. The intervals defined
bya < x < banda < x < bare written [a, b) and (a, b]. The infinite
interval [a, 00) consists of the points x such that x > a. Correspond-
ing interpretations are put on the expressions (a, 00), (—©, a),
(©, a]. Finally, (— 00, 0) is R.
Exercises 1(a)
1. Prove that
@ AUB=A=+A>B; Gi) ANB=A+ACB.
2. Prove that
(Bu C)n(Cu
A)n (AUB) = (BN O)u(CN
A)U (AN B).
3. Show that A—B = Af B’. Hence or otherwise prove that
@) (A—B)n(A—-C) = A-(BUC) = (A-B)-C;
Gi) (A—B)U(A—C) = A-(B0Q);
Gii) (A—C)n (B—C) = (An B)-C;
(iv) (A—C) U(B-C) = (AU B)-C.
4. Prove that
(@) (€A-B)nC= (AnC)-(BnO;
(i) (A-B)UC = (AUC)-(BUQ)=Ce= oa.
5. Prove that
@ (AUB)-B=A+ANB= Q;
(ii) (A-B)UB=A+A>B.
6. The symmetric difference A A Bof the sets A, Bis d
¢ ) ’ N fined as the set of el
in A or in B, but not in both. Show that ae
44 B= (A~B)U (B—A) = (AU B)—(4nB) = (AU B)n(4'UB).
Deduce that
@MASB=9+A=B; (ii) A’ AB=AAB,
7. Prove that
@AAB=BAA; (ii) AABAC=AA(BAO,.](https://image.slidesharecdn.com/22284738-250620074959-26290e84/85/A-Second-Course-In-Mathematical-Analysis-J-C-Burkill-H-Burkill-18-320.jpg)
![1.1] SETS AND NUMBERS 5
8. Prove that
@ AA BNC= (ANC)
A (BNO);
@) AA BUC=(AUQABUO=$CeH= oa;
ii) (4 A B)U(AA C) = (AU BUC) n(4’U BUC).
9. Show that AAB=CAD#AAC=BAD.
10. When Eis a finite set (i.e. one with finitely many elements), denote by |E| the
number of its elements.
Show that, if the sets A, B are finite, then
|A|+|B] = |4u Bl +]4nB}.
1.2. Ordered pairs and Cartesian products
A set with elements a, b, c, ... is often denoted by the symbol
only ae (1.21)
The notation calls for a number of comments.
(i) In such a listing of elements it is immaterial whether a particular
element appears once or several times. This convention is purely a
matter of convenience. For instance it allows us to denote the set of
roots of a complex quadratic equation by {a, b}, whether a + b or
a=b.
(ii) The order in which a, b, c, ... are written in {a, b, c, ...} has no
significance. For example {a, b} = {b, a}.
(iii) It is important to distinguish between the object a and the
set {a}, i.e. the set whose only member is a. Thus the set {@} has one
element, while @ has none.
The notation (1.21) has a useful variant. Given a set X, denote by
P(x) a statement, which is either true or false, about the element
x of X. The set of all x € X for which P(x) is true is written
{x € X|P(x)},
or simply {x|P(x)}
if the identity of the set X is clear from the context.
Illustrations
@ {xEZ|x?+1 = 0} = fi, —i;
(i) (ce Rx2+1 = 0} = 2;
ii) {xe R'la < x < 5} = [a, A).
The notion of a set does not involve any ordering among the
elements. For instance in the set {a, b}, a and b have the same status.](https://image.slidesharecdn.com/22284738-250620074959-26290e84/85/A-Second-Course-In-Mathematical-Analysis-J-C-Burkill-H-Burkill-19-320.jpg)
![6 SETS AND FUNCTIONS [1.2
The ordered pair (a, b) is the set {a, b} together with the ordering ‘first
a, then b’. Thus (a, ) and (6, a) are different unless a = b. (The
context will determine whether the expression (a, 5) stands for an
ordered pair or an open interval.) ;
The intuitive concept of an ordered pair (a, b) may be formalized
by the definition (ab) = tah, (a, BN,
(See exercise 1(5), 1.)
An ordered set (a, ..., @,) of any finite number n (> 1) of elements
is defined in a similar way.
Definition. Given the non-empty sets X4,...,Xn, their Cartesian
LOS, yoy (1.22)
is the collection of all ordered sets (x1, ...) Xn) such that
Xy CAG, 55 Xp, SC Aq
IfX, =... = X, = X, the set (1.22) is denoted by X"; X1is taken to
be X.
For n > 1, R® = R'x...x R! (n factors) is the set of all ordered
sets (x1, ..., X,) of real numbers. We define an interval in R” as a set
Ty Xsan MIs
where J,, ..., J, are intervals in R'. For example [a, b] x[c, d] is the
set of points (x, y) € R? such thata <x <bandc<y<d.
1.3. Functions
Let X, Y be two non-empty sets. A relation from X to Y is a subset
of Xx Y. Ifa relation
f, i.e. a subset of Xx Y, is such that, for every
x € X, there is one and only one member (x, y) of f, then fis said to be
a function on X to (or into) Y; we express this symbolically by writing
Six
= YF.
The set X is called the domain of f.
In some contexts the terms mapping, transformation, or operator
are often used as synonyms for function. Let (x, y) be an element of
the function f: X > Y. Then we say that y is the value of f at x (or
the image of x under
f) and we write y = f(x). If Lis any subset of X,
the subset of Y (ye Yly =f) and xe}
is called the image ofE under
/; it is denoted by f(E). The set f(X) is
called the range of f, Note that f(X) may be a proper subset of Y.](https://image.slidesharecdn.com/22284738-250620074959-26290e84/85/A-Second-Course-In-Mathematical-Analysis-J-C-Burkill-H-Burkill-20-320.jpg)
![1.3] FUNCTIONS 7
If a function fhas its domain in X and its range in Y, we say that
Sis a function from X to Y. A function from R! to R! will be called a
real function, a function from Z to Z a complex function.
Illustrations
(i) We obtain a function on X to Y whenever we have a rule for assigning a
single value y in Y to each x of X. Thus the equation x?+y? = 1 defines a
function on R! to R! whose range is (— 09, 1].
Gi) The equation x*+y? = 1 defines a relation but not a function on [—1, 1]
to [—1, 1] because, whenever — 1 < x < 1, it is satisfied by two values of y.
Gii) The logarithmic function is the function on (0, 00) to R' which assigns
the value log x to the positive number x.
(iv) The sequence ay, a2, as, ... is a function whose domain is the set of positive
integers exhibited as suffixes.
More generally, given two arbitrary non-empty sets A, Janda function on Jto
A, we sometimes find it convenient to write the image of a member / of J in the
form a;,.
(v) Denote by © the set of all real functions ¢ defined by an equation of the
form
G@) = axt+fh (xe R),
where a, f are real numbers. A function f: R* > ® is obtained by assigning to
each point (a, b) € R* the function ¢,,, given by ¢.,(x) = ax+b (xe R%).
Let
f be a function on X into Y. If f(X) = Y, ie. if every ye Yis
the image under fof at least one x € X, then we say that fis surjective,
or that fmaps X onto Y.
If different elements of X have different images, i.e. if f(x) = f(x’)
implies x = x’, then fis called injective.
If f is both surjective and injective, so that every value y in Y is
assumed once and once only, then fis said to be bijective. A bijective
function is also called a bijection; it effects a one-to-one correspondence
between its domain and range.
Consider again the illustrations above.
(i) The function maps R! onto (—, 1]; it is not injective.
(iii) The logarithmic function is a bijection on (0, <0) to R*.
(v) This function is a bijection on R* to ®.
Inverse functions. Suppose that the function f has domain X and
range Y. Tof, which is a subset of X x Y, there corresponds the sub-
f Yx X defined
Sine GNIGIER: (31)
Plainly this is a function with domain Y and range X if and only if f
is bijective. When f is bijective, (1.31) is called the inverse Function
of f and is denoted by f—. For instance the logarithmic function has](https://image.slidesharecdn.com/22284738-250620074959-26290e84/85/A-Second-Course-In-Mathematical-Analysis-J-C-Burkill-H-Burkill-21-320.jpg)
![1.3] FUNCTIONS 9
Theorem 1.31,
(i) If the function f: X + Y is surjective and if there is a function
&: Y>X such that
go f = ix, then
f is bijective and g = f>.
(ii) If the function f: X + Yis bijective and if the function g: Y > X
is such that
fo g = iy, theng = f-.
Proof.
@ IEf@) = f(x), then
x = g{f(x)} = aff} = x’
and so fis injective as well as surjective. We now have
g = goiy = go(fof) = (gof)oft = ixoft =f.
(i) g = ixog = (flof)og = fto(fog) =froiy =f". |
In theorem 1.31 (ii) the bijectiveness of f is not a consequence of
surjectiveness, as it is in part (i). For instance let X = {a, b} (a + b),
Y = {c} and let f, g be defined by f(a) = f(b) = c, g(c) = a. Thenf
is surjective and fo g = iy, but fis not bijective.
If the functions f: ¥ > Y, g: Y > W are bijective, then go/f:
X — W is bijective. It follows from theorem 1.31, or it may be
proved directly, that (gof)3 = frog.
Restriction and extension of functions. Suppose that X, > X, and
that f: X, > Y, g: X, > Y are functions such that f(x) = g(x) for
all x € X,. We then say that g is the restriction of f to X, and that fis
an extension of g to X,. For example let f,: R! > R' and f,: R! > R?
PE RTEU HN T A) p fts) — [x] Ae RY,
The function g: [0, 00) > R! defined by
a(x) =x (x 20)
is the restriction to [0, 00) of both f; and f2; and f,, f, are both
extensions of g to Ri.
Exercises 1(5)
1. Show that {{a}, {a, b}} = {{c}, {c, d}} if and only if a = cand b = d.
2. Prove that
@ (AUB)xC= (Ax C)uU (Bx),
(ii) (An B)x C = (Ax C)n (Bx),
(iii) (A—B)x C = (Ax C)-(Bx C).](https://image.slidesharecdn.com/22284738-250620074959-26290e84/85/A-Second-Course-In-Mathematical-Analysis-J-C-Burkill-H-Burkill-23-320.jpg)
![1.4] SIMILARITY OF SETS 11
It is easy to see that similarity has the following three properties:
(i) A ~ A (reflexivity);
(ii) if A ~ B, then B ~ A (symmetry);
(iii) if A ~ Band B ~ C, then A ~ C (transitivity).
If we confine ourselves to a given collection @ of sets, then simi-
larity is a relation from @ to @ and, in view of (i)-(iii) above, it is an
equivalence relation. The reader has probably met such relations in
other branches of mathematics, particularly algebra.
Finite sets are similar if and only if they have the same number of
elements. This implies, in particular, that a finite set cannot be
similar to a proper subset of itself. We shall see that infinite sets
always contain proper subset: hich they are similar (theorem 1.42,
corollary). A simple example is the following. Let A be the set of all
integers and let B be the set of all even integers. Then A ~ B, since
the function given by y = 2x (xe A, ye B) establishes the required
one-to-one correspondence.
A set which is similar to the set {1, 2, ...} of positive integers is said
ee ee Such a set is therefore one
that can be arranged as a sequence. A set is called countable if it is
either finite or countably infinite.
We now show that there are infinite sets which are not countable.
Theorem 1.41. The set of real numbers in the interval (0, 1) is not
countable.
Proof. Suppose that the real numbers in the interval (0, 1) do
form a countable set. This set is certainly infinite and can therefore
be arranged as a sequence
Ay, Az, Ag, «+++
Now represent each number a, as an infinite decimal
An = 0.0n1%ne%ng +
in which recurring 9’s are not used. Such a representation is unique.
Define h f when @,, + 1,
Then b=0.f, hfs -..
is a real number between 0 and 1 which differs from a, in the nth
decimal place. Hence b is not one of the numbers @;, ds, ds, ... and
this contradicts the original assumption that all real numbers in the
interval (0, 1) appear in the sequence (a,). |
2 when«,, = 1.](https://image.slidesharecdn.com/22284738-250620074959-26290e84/85/A-Second-Course-In-Mathematical-Analysis-J-C-Burkill-H-Burkill-25-320.jpg)
![1.4] SIMILARITY OF SETS 13
Proof. The elements of P%, all of which appear in the array
(1, D, G, 2), C, 3).
(2, 1), 2; 2), (2, 3), +
(3, 1), G, 2), G, 3), -.-,
may be arranged as a sequence in several simple ways. Perhaps the
simplest is the enumeration by diagonals:
G0); 2, (2); GD, 22), 0,3); «3
1), @-1, 2)... Asn); |
Exercise. Show that the set P" (n > 2) of all ordered sets (p,, ..., Dn) of n positive
integers is countable.
pee
Theorem 1.45. The union of a countable collection of countable sets
is countable.
Proof. Let {A, Ao, ...} be a finite or countably infinite collection
of countable sets. If
=i
B= A, B,=An—U Ay (2 > 1),
k=1
then the B, are mutually disjoint and
UA, =UB, =C,
n n
say. (The non-committal notation U is used since the number of sets
n
may be finite or infinite.) By theorem 1.43 each of the sets B,, is
countable. Therefore, if B,, is not empty, its elements may be arranged
in a possibly terminating sequence
Daas Bras «+++
Now let S be the subset of P? which contains a pair (p, q) if and only
if B, is either infinite or has at least gmembers. By theorems 1.43 and
1.44, S is countable. Since there is an obvious one-to-one correspon-
dence between S and C, it follows that C is countable. |
Theorem 1.46.
(i) The set of all rational numbers is countable.
(ii) The set of all irrational numbers is not countable.](https://image.slidesharecdn.com/22284738-250620074959-26290e84/85/A-Second-Course-In-Mathematical-Analysis-J-C-Burkill-H-Burkill-27-320.jpg)
![1.4] NOTES 15
8. A set E of real numbers is such that every series
=
2 Xe
n=1
whose terms are distinct elements of E, converges. Show that E is countable.
9. Given a set A, denote by “4 the set of subsets of 4 (including @ and A).
Show that
@ if A has n elements, then 4 has 2” elements;
(i) if A is countably infinite, then %4 ~ R};
oe if A is any set, then A + %4 (though clearly A is similar to a subset of
“A)»
10. Show that any collection of disjoint open intervals of R' is countable.
Deduce that, if J is any interval in R' and the function f: J > R* is monotonic,
then the set of points of discontinuity of f is countable.
(if ce¢J and c is not an end point, then
fet) = lim f(x), f(e—) = lim f(x)
z>c+ ze
exist; see C1, 32.)
NOTES ON CHAPTER 1
§1.1. Axiomatic set theory. The intuitive notion of a set, though perfectly
adequate for everyday use by analysts, is, in fact, self-contradictory. To demon-
strate this we describe the famous Russell paradox. A set may or may not be a
member of itself: most sets are not and we call these normal; an example of a set
which is a member of itself is the set consisting of all infinite sets (for clearly
there are infinitely many of these). Consider the set N of all normal sets. Is N
normal or not? If N is normal, then, by definition of normality, N ¢ N; but, by
definition of N, a set which does not belong to N is not normal and we have a
contradiction. The supposition that N is not normal is equally untenable. The
mathematician’s way out of this dilemma is to construct an axiomatic system
designed to mimic the desirable features of the intuitive approach and to avoid
its pitfalls. The axiomatic theory of sets is described in Theory of Sets and
Transfinite Numbers by B. Rotman and G. T. Kneebone. A more informal atti-
tude is adopted by P. R. Halmos in Naive Set Theory.
The theory of sets is the starting point of the monumental Eléments de
Mathématique, which is designed to present the whole of pure mathematics in
strictly logical order. This work, by a group of French mathematicians writing
under the collective name of N. Bourbaki, has been appearing in sections since
1939 and is not yet (in 1969) complete.
The real number system. The set R’ of real numbers, together with the operations
of addition and multiplication and an ordering relation, is an ordered field: it
satisfies the following six algebraic and three ordinal axioms.
A1. To every pair of real numbers a, b correspond a real number a+, the
sum of a and b, and a real number ab, the product of a and b.
A2. a+b = b+a; ab = ba.
A3. (at+b)+c = a+(b+0); (ab)e = abe).
A4. There are real numbers 0, 1 such that
O+a=a and la=a
for all real numbers a.](https://image.slidesharecdn.com/22284738-250620074959-26290e84/85/A-Second-Course-In-Mathematical-Analysis-J-C-Burkill-H-Burkill-29-320.jpg)
![18 METRIC SPACES [eeu
Gi) On R? the usual metric is given by
p(x, ») = |x-yI-
The usual metric on R" is defined as follows. If
x= Gow, PH One eke
px, ») = M@a—
yi)? +... +n In). (2.11)
The case n = 1 gives the earlier definition of the usual metric on R’.
Plainly p satisfies M1 and M2. Putting a; = x:—yi, b; = y:—Z; we see that
M3 is equivalent to
n t n +
(a a+b) < ( > a) +(% 7) if (2.12)
= i=1 i=
By squaring both sides of (2.12) we obtain yet another equivalent form, namely
n n t/n +
Sais (3 a) (3 vi) : 2.13)
i=1 i=1 i=
This now follows from the fact that, for all real &,
3 (a€+b)? = A?+2HE+B > 0,
i=1
so that H? < AB,
n n n
where A= a, H= Yad, B= > i.
i=l i=l =
Therefore p, defined by (2.11) is a metric.
The metric space (R", p), where p is the usual metric, is called n-dimensional
Euclidean space.
(iii) On Z, the complex plane, the usual metric is of the same form as that
on R*. If z = x+iy and z’ = x’+iy’ (x, y, x’, y’ real),
PG, 2’) = |z—z'| = Me—-xP+0-y93.
In fact, when equipped with their usual metrics, Z and R? are essentially the same
metric spaces in which only the notations differ.
(iv) Let Bla, 6] be the set of bounded real functions on the interval [a, b] and,
for f, Bia, b), |
Thee Bahl ope) = sup |f)—ea)-
asr<bh
Clearly p has the properties M1 and M2. M3 follows from the inequalities
sup |A(x)+ ¥()| < sup (|$@)|+ |¥@)|) < sup |4(2)|+sup |¥Q)].
(v) Let S be the space of sequences x = (x, Xa, ...) of complex numbers such
that Xx, converges absolutely. For x, y € S put
Ea
PY) = Ds [Xn—Yal-
n=1
It is easy to show that p is a metric on S.
Let (X, p) be a metric space and let Y be a subset of XY. If o is thicy
restriction ofp to Yx Y, ie. o(x, y) = p(x, y) for all x, ye Y, then.](https://image.slidesharecdn.com/22284738-250620074959-26290e84/85/A-Second-Course-In-Mathematical-Analysis-J-C-Burkill-H-Burkill-32-320.jpg)
![2.1] METRICS 19
¢ is a metric on Y; it is said to be the metric induced by (X, p) on Y.
The metric space (Y, 0) is called a (metric) subspace of (X, p). For
instance let R? have its usual metric p given by
P(x, Y) = V{Ci— yd? + 2 — Ya)? + %s—Ys)"}-
If Y is the set of points x € R* with x; = c, the metric o induced on
Yis given by — o(x, y) = V{Ca-y)*+ (yo
In future, unless there is a statement to the contrary, the metrics
on R” and Z will be taken to be the usual metrics; and subsets of R”
and Z will be taken to have the metrics induced by the usual ones.
Sequences of points. The following definition extends the notion of
convergence of sequences of real numbers to sequences of points in
a metric space.
Definition. The sequence (x,) of points in the metric space (X, p) is
said to converge to the point x(€ X) if
P(Xn, xX) +0 as n>. (2.14)
We write, as in classical analysis, X, > x as n> or lim X, = x.
n>
Since p(x, y) is a real number, we are able to use the concept of
convergence in R!. We could replace (2.14) by the phrase ‘given
e > 0, there is an m such that p(x, x) < € for n > m’. This form
of the definition reduces to the familiar definition of a convergent
sequence of real numbers when (X, p) is R' with its usual metric.
Exercise. Prove that 4 sequence cannot converge to two distinct limits.
Illustrations
y RI = n 2n* )
@ In let Xn = nti? n—2)°
Then, as n > 00, Xn => (8, 2);
since PGrm 2) = ee aye Seal me
Gi) In Bil, 2] with pg) = sup _|f(x)—g(), let
f, be given by
1<a2<2
f(x) = (+x (1 < x < 2).
Then Sih
where GQ) —* =x = 2),
For, when 1 < x < 2,
0 < fl)
—f 0) = xO?
+ I)" — x < x(2¥8—1) < 2(2""—-1)
and so Pat) < 224"-1)>0 as n>.](https://image.slidesharecdn.com/22284738-250620074959-26290e84/85/A-Second-Course-In-Mathematical-Analysis-J-C-Burkill-H-Burkill-33-320.jpg)
![2.1] METRICS 21
8. On R®, o and 7 are defined by
a(x, y) = max (|x:—yi], |x2—yal), 7,9) = [x1 — yal +] %2—yal
Show that o and 7 are metrics which are equivalent to one another and to the
usual metric p.
Compare the sets of pointsx in R® given by p(x, a) < 1,0(x, a) < 1,7(x,a) <1
Tespectively.
9. Denote by ¢ the discrete metric on R! (p. 17, illustration (i). Prove that o and
the usual metric p on R! are not equivalent.
10. Let C be the set of real functions continuous on the interval [0, 1] and define
ene m +)
Laparaet phe) = sup [fe)-so,
0<e<1
‘1 +
otha) = ([ 1rco-eta)* dx) .
‘1
rhe)= | 1f)-et0) ax
(A function continuous on a closed interval is bounded and integrable.) Prove
that o and 7 are metrics on C. Show also that, for all f, g € C,
PL, 8) > OC,8) > TU, 8).
For n = 1, 2, ..., the functions
f,, g, € C are defined by
l-nx O<x<7nr), n(i—nx) (0<x <n),
(x) = { &nlx) = {
) i el) 0 rs <ix.<, 1);
Prove that the sequence (f,,) converges in (C, o) and in (C, 7), but not in (C, p);
and that (g,,) converges in (C,7), but not in (C, p) or in (C, ¢).
11. Let (X4, p,), ..., (Xn, Pn) be any metric spaces. Verify that metrics on
X,x ... x X, are given by
o(x, y) = max {P,(%1, Va), «++» Pn(Xns Yn)},
Ox, ¥) = VPI, Vi) t+... + PA Xn» Ynd},
OX, ¥) = PilXr, V+... +Pnl%ns Yn)s
where x = (x1, ..-5 Xn), ¥ = Oy -++5 Yn) and xi, ¥i € X; @ = 1, ..., 2). Show also
that 0, 72, 03 are all equivalent.
2.2. Norms
The theory of vector spaces is partly algebraic and partly analytical.
We need only the most basic algebraic ideas and these are, no doubt,
known to the reader. But, largely to fix notation and terminology,
we formally define a vector space.
Definition. Let V be a non-empty set and suppose that (i) to any two
elements x, y of V there corresponds an element in V, called their sum,](https://image.slidesharecdn.com/22284738-250620074959-26290e84/85/A-Second-Course-In-Mathematical-Analysis-J-C-Burkill-H-Burkill-35-320.jpg)
![22 METRIC SPACES [2.2
which is denoted by x+y; and (ii) to any real number « and any x in V
there corresponds an element in V, called the scalar product of « and
x, which is denoted by ax. (Thus addition is a function on V x V into
Vand scalar multiplication is a function on R'x V into V.) Suppose
also that the following conditions are satisfied.
() x+y = y+;
(2) x+(y+z) = (x+y)+23
(3) there is an element 0 € V (the zero element) such that x+0 = x
for allx eV;
(4) to every element x€V there corresponds an element —xeV
such that x+(—x) = 9;
(5) a(Bx) = (@B)x;
(©) 1x =x;
(7) (@+A)x = ax+ Bx;
(8) a(x+y) = axtay.
Then V (together with the operations of addition and scalar multi-
plication defined in V) is called a vector (or linear) space over R', or a
real vector space.
If the scalar product is formed with complex numbers « and the same
properties (1)-(8) hold, then V is said to be a vector space over Z, or a
complex vector space.
The simplest example of a real vector space is R” with addition
and scalar multiplication defined by
(ay +25 Xn) +O, ---s Ra) = + Vv «+s Xn tPn),
A(X, -++9 Xn) = (WX, «+, WXp)-
A more sophisticated example is Bla, b]; here f+g and «fare defined
by
F+9® =f/@%)+s@), CO=7%0) @<x< bd).
We shall assume the elementary algebraic properties of vector
spaces. Some of these are given in exercise 2(b), 1.
Definition. Let V be a vector space. The real valued function || .|| on V
is called a norm on V, and V is called a normed vector space, if
N1. ||x|| > 0 for all
x € V and ||x|| = 0 if and only if x = 0;
N2. |x+y| < ||| +ly] for all x, ye V;
N3. ||ax|| = |e|||x| for every number « and all x € V.
It is easy to see that, in a normed vector space, a metric p may be
defined by
A(x, y) = |]x—y]-](https://image.slidesharecdn.com/22284738-250620074959-26290e84/85/A-Second-Course-In-Mathematical-Analysis-J-C-Burkill-H-Burkill-36-320.jpg)
![2.2] NORMS 23
In fact, the metrics in illustrations (ii)-(v) of §2.1 all arise from
norms. For instance in R” the usual norm is given by
[|x] = V@i+... +27)
and in B[a, b] |fl = sup IfC0)].
arc
However, a metric on a vector space need not be derivable from a
norm (see exercise 2(6), 2).
More specialized than normed vector spaces are inner product
Spaces.
Definition. Let V be a complex normed vector space and suppose that,
with every ordered pair (x,y) of elements in V there is associated a
complex number, denoted by x.y, and that the following conditions are
satisfied.
Il. x.y = y.x for all x,y eV;
12. (ax+fy).z = a(x.z)+f(y.z) for all numbers a, 8 and all
x,y, ZEV;
13. x.x = |x|? for all xe V.
Then x.y is called the inner product of x and y and V is called a
complex inner product space.
If V is areal normed vector space, x.y is real and 11-13 hold, then V
is called a real inner product space. Note that, in this case, I1 reduces
toe ya yer:
It is shown in exercise 2(b), 4 that no more than one inner product
can be associated with a given norm.
Once again R” provides the most obvious illustration. The inner
product for the usual norm is defined by
XV = Mite
+ kane
The inequality (2.13) can now be written as
x.y < |x] [vl
and is a particular case of the following result.
Theorem 2.21. (Cauchy’s inequality.) If V is a real or complex inner
product space, then, for all x, y € V,
Equality holds if and only if one of x, y isa scalar multiple of the other.](https://image.slidesharecdn.com/22284738-250620074959-26290e84/85/A-Second-Course-In-Mathematical-Analysis-J-C-Burkill-H-Burkill-37-320.jpg)
![24 METRIC SPACES [2.2
Proof. Since 0. = 0 (see exercise 2(b), 6), we may assume that
x + 0. Take yee
Then 0 < ||Ax+yl/? = (Ax+y).(Ax+y)
= ew oben elie
— le¥lP ya 9 1:
— Fak bp 2te
Moreover ||£x+yl| = 0, ie. Ax+y = 6, if and only if y = ax for
some number «. |
Exercises 2(5)
1. Show that the following identities hold in a vector space:
(i) Ox = 0; (ii) (-I)x = —x; (iii) 20 = 0;
(iv) if x—y denotes x+(—y), then
Ey) a ee YX
2. Prove that the discrete metric 7 on R? (see illustration (i) of § 2.1) cannot be
derived from a norm.
3. Two norms on a vector space V are equivalent if they give rise to equivalent
metrics, Prove that the norms |.|, and |], are equivalent if and only if there
exist strictly positive constants A, 4 such that
Axl, < Ixle < allt,
for all x € V. (Contrast exercise 2(a), 7.)
4. Show that if, in a normed vector space, an inner product exists, then it is
unique.
5. The space T consists of the sequences x = (x,, X2, ...) of complex numbers
such that &|x,|? converges. Verify that p given by
oss) = ( beam aalt) Ge yeT)
is a metric on 7; that the metric arises from a norm; and that to this norm there
corresponds the inner product
©
X.Y = YD XnTne
n=l
6. Show that in an inner product space x.0=0.x=0.
7. Prove that, if V is an inner product space, then, for all x, ye V, z
[x+y]?+ xy]? = 2(1x|?+
| 719-
Hence give an example of a normed vector space which is not an inner product
space.](https://image.slidesharecdn.com/22284738-250620074959-26290e84/85/A-Second-Course-In-Mathematical-Analysis-J-C-Burkill-H-Burkill-38-320.jpg)
![2.2] NORMS 25
8. The sequences (@n), (Yn) in the normed vector space V converge to x,y
Tespectively, and the numerical sequences (@n), (2) converge to a, 2 respectively.
Show that
%mXn+ Bn¥n > &x+ By.
Prove also that, if V possesses an inner product, then
Xn-Yn
> X.Y.
9. Alternative definition of inner product space. Let V be a complex (real) vector
space and suppose that, with every ordered pair (x, y) of elements in V, there is
associated a complex (real) number x.y such that I1, I2 are satisfied and
13’. for all xe V, x.x > O and x.x = 0 if and only ifx = 0.
Show that ||x|| = ./(x.x) is a norm on V.
2.3. Open and closed sets
In this section we consider subsets of a fixed metric space (X, p).
We shall use the symbol X for the metric space as well as the set
without reiterating that p is specified.
Let ae X and let r be a positive real number. The set of points x
such that p(a, x) < ris called the open ball with centre a and radius r;
we shall denote it by B(a;r). In R', B(a;r) is the open interval
(a—r, a+r). In R? and in Z an open ball is usually called an open
disc.
—
Definition. Given a subset E of X (which may be X itself), the point
cé€X is said to be a limit point of E if, for every € > 0, there is an
x €Esuch that 0<p(ce,x) <6,
i.e. B(c; €)—{c} contains a point of E.
A limit point c of E can also be defined by either of the following
two conditions: fj
(i) Every open ball B(c; ¢) contains infinitely many points of E.
(ii) There is a sequence of points x, ¢ such that x, +c and
X, > casn>o.
It is clear that a point c with properties (i) or (ii) is a limit point ofE.
Conversely, if c is a limit point of E, then, for every n, there is a point
X, € E such that 0 < p(c, x) < In
and so (ii) is satisfied. Also (ii) implies (i).
The conditions (i) or (ii) show that a finite set cannot have a limit
oint. An infinite set may or may not have a limit point. For example
the subset {1, 2, 3, ...} of R! has no limit point.
It is important to note that a limit point of a set Z may or may not
belong
to £. For instance if, in R', E is the interval (0, 1), then every
point in the interval [0, 1] is a limit point of E.](https://image.slidesharecdn.com/22284738-250620074959-26290e84/85/A-Second-Course-In-Mathematical-Analysis-J-C-Burkill-H-Burkill-39-320.jpg)
![is
Nu open" tera SSS SRC IGAsal) eGo Ras
26 METRIC SPACES [2.3
A point c of E is sai isolated poi) E if it is not
point of E. Thus, if c is an isolated point of £, there isa é > Osuch
laaroagat Somes . .
that B(c; 6) contains no point of £ other than c.
Definition. Given a set E < X, the point c € Eis said to be an interior
point of E if there is a 8 > 0 such that B(c; 8) © E.
Illustrations
Ai All points of an open ball B(a; r) are interior points of B(a; r). For suppose
that c € B(a; r), so that p(a, c) = s < r. If 6 = r—s and p(c, x) < 4, then
pa, x) < pla,c)+plc, x) < st+d=r
and so x€B(a;r). Therefore B(c; 6) < B(a;r), i.e. c is an interior point of
Ba; r).
Gi) X = Ri.
Let E = [0, 1). The set of limit points of E is [0,1]; the set of interior points
of E is (0, 1).
—~TLet
FE be the set of rational points of Rt. Every point of R* is a limit point of E.
E has no interior points.
Gi) X = Z.
Let E be the annulus consisting of the points z such that 1 < |z| < 2. The set
of limit points is the set of z such that 1 < |z| < 2. The set of interior points
is E.
Let E be the set of points x+iy such that y = —1 and 0 < x < 1. The set of
limit points is E. There are no interior points.
When the space X has no isolated points (as is the case with R”
an then an interior point of a subset E of X is also a
of E. But if X is, for instance, the space consisting of
allth
(andp is the usual metric in R¥), then every subset E of X is without
wits limit points although all the points of E are interior points of E.
HS. ie
—— Definition. (i) A set F in X is said to be closed if F contains all its
limit points.
(ii) A set G in X is said to be open if every point of G is an interior
point of G.
a Y
Illustrations
@ Every open ball in a metric space is open. A set {x|p(a, x) < r} is closed
(see exercise 2(c),
2) and is
consequently called the closed ball with centre a and
radius r. = - 7
Any set without limit points is closed; in particular, every finite set is closed.
V/ (ii) In R", every open interval is open and every closed interval is closed.
o/ (ii) In R', the interval [0, 1) is neither open nor closed; the set of rational
points is neither open nor closed,
(iv) In Z, the annulus defined by 1 < |z| < 2 is open; the real axis (im z = 0)
is closed; the set of points z = (1+i)/n(m = 1, 2, ...) is neither open nor closed.](https://image.slidesharecdn.com/22284738-250620074959-26290e84/85/A-Second-Course-In-Mathematical-Analysis-J-C-Burkill-H-Burkill-40-320.jpg)
![2.3] OPEN AND CLOSED SETS ZT
(v) Let (Ca, 5], p) be the metric space of real functions continuous (and so
bounded) on the interval [a, 5], in which
PG 8) = oon |f@)—-g@)|-
a<r<
Then the set E, of f such that
inf f(x) >0
a<x<b
is open and the set E, of f such that
f@=1
is closed.
(vi) In any metric space the empty set and the whole space are both open
closed. In R” and
Z these are the only sets that are both open and closed (theorem
-$:37, corollary). In an arbitrary metric space there may be others. For instance
in the space of integers (with the usual metric of R) every set is both open and
closed. The question is taken up again in §3.3.
(Wii) The property of being open or closed (or neither) is not intrinsic to a
given set £; it is relative to t i i subset
metric p wil i i . For instance [0, 1) is neither open nor closed
as a subset of Rt equipped with the usual metric, but [0, 1) is closed as a subset
of (—1, 1) with the same metric; and as a subset of (R!, 7), where is the discrete
metric, [0, 1) is both open and closed.
A convenient characterization of closed sets is embodied in the
following result.
Theorem 2.31. A necessary and sufficient condition for the set F to be
closed is that, whenever (x,,) is a convergent sequence of points in F,
lim x, € F.
no
Proof. First suppose that F is closed and that x, — x, where
X, € F for every n. If x,, = x for all sufficiently large n, then, a fortiori,
x € F; otherwise x is a limit point of F and again xe F.
Next, let F have the property described in the theorem. If F has no
limit points, then F is closed. If F has a limit point x, then there is a
sequence (x,) of points in F such that x, > x, and so xe F. |
Theorem 2.32. (i) The set G in X is open if and only if G’ (= X—G)
is closed.
(ii) The set F in X is closed if and only if F’ (= X—F) is open.
Proof. We need only show that (a) if G is open, then G’ is closed
and (6) if Fis closed, then F’ is open.
(a) Let G be open. If G’ has no limit points, then G’ is closed. If G’
has a limit point c, then every open ball B(c; ¢) contains points of G’.
Hence c cannot be an interior point of G and, since G is open, c ¢ G.
Thus ce G’. We have therefore shown that G’ contains all its limit
points.](https://image.slidesharecdn.com/22284738-250620074959-26290e84/85/A-Second-Course-In-Mathematical-Analysis-J-C-Burkill-H-Burkill-41-320.jpg)
![30 METRIC SPACES (2.3
It may be shown directly that £ is closed. The result also follows
from the identity (BY = (EY
which is easily proved. (See exercise 2(c), 7.) |
Open and closed sets could also have been defined in terms of
interiors and closures: G is open if and only if G° = G; Fis closed if
and only if F = F.
Definition. A frontier point of a set E in X is a point c such that
every open ball B(c; €) contains at least one point of E and at least one
point of E'. The set of all frontier points of E is called the frontier of E
and is denoted by fr E.
Clearly E and E’ have the same frontier.
Since a frontier point of E is a point of E which is a limit point of
E’ ora point of E’ which is a limit point of Z, it follows that
fr E = E-E°. (2.31)
Illustrations
(i) Let E be the interval [0, 1) in R*. Then E° = (0, 1), E = [0, 1] and frE =
{0, 1}.
(ii) Let (X,) be the space of integers with the metric induced by R. If
E= B¢c; 1) = {c}, then E° = E = E. Note that E is not the closed ball
{x|p(x, c) < 1} = {e-1, c, c+ 1}.
Exercises 2(c)
1. Show that every infinite set Y may be equipped with a metric p which is such
that X has a limit point in CX, p).
2. Prove that, in any metric space (X, p), the closed ball {x|p(a, x) < r}is closed.
3. Prove that a bounded, closed set of real numbers contains its supremum and
infimum. .
4. Let E be an arbitrary set in a metric space. Show that the set E* of limit points
of E is closed.
5. Show that in R" with the usual metric any collection of disjoint open sets is
countable. Is this true for an arbitrary metric space?
6. (i) Show that E° is the union of all open sets contained in E and so is the
‘largest’ open set contained in E.
(ii) Show that E is the intersection of all closed sets containing E and so is the
‘smallest’ closed set containing E.
7. Prove that, for any subset E of X,
E°ufr
Ev (E)° = X
and @y =):
(The set (E’)° is called the exterior of E.)](https://image.slidesharecdn.com/22284738-250620074959-26290e84/85/A-Second-Course-In-Mathematical-Analysis-J-C-Burkill-H-Burkill-44-320.jpg)
![2.3] NOTES 31
8. Prove that frE= EnE’
and that fr E is closed.
9. (i) Prove that Bice Ub. =hy UU Be
but that, for an infinite collection & of sets E, generally the relation
UE >UE
Ecé Ecé
only holds.
(ii) Prove that, for any collection & of sets E,
NES EF
Ecé Ecé
Construct an example to show that identity need not hold even for a finite
collection.
Gii) Deduce the results for interiors corresponding to (i) and (ii).
10. The metric spaces (X, p), (X, 7) have equivalent metrics. Show that every
subset E of X has the same limit points and interior points in (X, p) and CX, c).
11. If EZ is a set in a metric space and the set D is such that
DSH S-D:
then D is said to be dense in E. (For instance the set of rational numbers is dense
in R'.) Show that, if C is dense in D and D is dense in £, then C is dense in E.
NOTES ON CHAPTER 2
§2.1. The idea of a metric space is due to M. Fréchet. In his doctoral thesis,
published in 1906, he examined various sets of axioms for a metric. The set
M1-—M:S3 is the one which experience showed to be the most fruitful.
The inequality (2.12), which is equivalent to the triangle inequality for the
usual metric on R”, is a particular case of Minkowski’s inequality
(3, oro)” < (Sa) (Se)
in which a,, b; > 0 (i = 1, ... n) andr > 1. (In (2.12) the a;, b; may be negative,
but the step from non-negative a;, b; to arbitrary ones is trivial.) The inequality
(2.13), which we call Cauchy’s inequality in theorem 2.21, is also known as
Schwarz’s inequality (or even the Cauchy-Schwarz inequality). It is a particular
case of Hélder’s inequality.
n n p/n 1/q
E ab, < (3 at) (3) :
i=l i=1 i=1
where a;,b; > 0 (Gi =1,..-,”), p> 1 and p++q = 1. Both Hdlder’s and
Minkowski’s inequalities may be generalized in a variety of ways. (See Hardy,
Littlewood and Pélya, Inequalities.)](https://image.slidesharecdn.com/22284738-250620074959-26290e84/85/A-Second-Course-In-Mathematical-Analysis-J-C-Burkill-H-Burkill-45-320.jpg)
![3.1] LIMITS 33
Gi) X = R*, FE = R*-{(0, 0)}, Y= R.
Let g, be given by Ga x
HOY) = ae
x?
Then lex | = xl eas lal
and lim __g,(x, y) = 0.
, (x, v)>(0, 0) .
Let gy be given by als) ==
ey
Then te paar 4 B(x, ¥) does not exist. For, if y _=
mx,we have (for all x + 0),
Be 1
82%, mx) = a
+ meet +m?
and so g, takes all values between 0 and 1 in any annulus 0 < x?+y? < 62, ew
(iii) Let E - X = ¥ = C(O, 1] (the set of real functions continuous on [0, 1]).
The metric p is, as is usual on C[0, 1], given by
Pb. $2) = sup |6@)—¢,(0.
The function A from C[0, 1] to C[0, 1] is given by A(¢) = ¥, where
Ho = [$00 ar.
If, for instance, ¢» is given by ¢o(t) = #2, then lim A(¢) = Wo, where yro(t) = 42°.
od.
For, if p($o, ¢) < ¢, then, for0 < ¢ < 1,
rea fo ar| = | [eran ar|
< [ir-senlar <e
0
ho-YO| =
and therefore p(yro, y) < €.
A sequence 4, ds, a; ... may be regarded as a function on the set
{1, 4, 4, ...}. It follows that the notion of a sequential limit may
be expressed by means of the notion of a functional limit. What is
less obvious, and more interesting, is that the opposite statement is
also true.
Theorem 3.1. Let (X,p) and (Y,0) be metric spaces. Let f be a
function with domain E < X and with range in Y and suppose that xo
is a limit point of E. Then f(x) > yo as x > Xp if and only if, for every
sequence (Xp) in E—{xo} such that x, > Xo, f(Xn) > Yo asin > 00.
Proof. (i) Suppose that f(x) > Vo as x > Xo. Then, given ¢ > 0,
there is a 6 > 0 such that
o(f(x), Yo) < € whenever x € E and 0 < p(x, x9) < 9.
Bsc](https://image.slidesharecdn.com/22284738-250620074959-26290e84/85/A-Second-Course-In-Mathematical-Analysis-J-C-Burkill-H-Burkill-47-320.jpg)
![34 CONTINUOUS FUNCTIONS ON METRIC SPACES [3.1
Now, if (x,,) is any sequence in E—{xo} such that x, > Xo, there isan
integer m such that
0 < p(Xn, Xo) < & for n> Mm.
Therefore o(f(Xn)s Yo) < € for n> Mm,
ne S(%n) > Yo AS n> OO.
(ii) Suppose that, whenever (x,) is in E—{x} and x, > Xo,
F(%n) > Yo-
If f(x) +> yo aS X > Xo, then there is an ¢ > O with the property
that there is no 6 > 0 such that o( f(x), yo) < € whenever x € E and
0 < p(x, xo) < 6. Therefore, given any n, there is an x, € E such that
0 < (Xn, Xa) < In and o(f(%q), Yo) > &
Thus there is a sequence (x,) in E—{xo} such that x, > x) and
F(%n) +> Yo. This contradicts our original hypothesis and so f(x) > yo
as X > Xp. |
For functions with values in a normed vector space V there is, as
might be expected, an algebra of limits. Let f; g be such functions
defined on a subset E of X. Also suppose that E has a limit point x9
— L]) > Yoo B(x) > 2% aS X> Xp.
When ¢, are real or complex valued functions on E and
d(xa>a, Yx>h as x>X,
then PODS) + VOQ)B(x) > 2Y0 + Azo. (3.11)
If V possesses an inner product, then also
F(X) .8(%) > Yo-Z05 (3.12)
and if Vis R? or Z, F)/g(X) > yolZ0 (3.13)
provided that z) + 0.
These results may be proved directly or they may be deduced, by
means of theorem 3,1, from the corresponding results on sequences
(exercise 2(b), 8).
Exercises 3(a)
1. The real valued function fon R?—{(0, 0)} is defined by
x*y
fy) = Taya
Show that lim f(x, y) does not exist.
(@, v)—>(0, 0) va](https://image.slidesharecdn.com/22284738-250620074959-26290e84/85/A-Second-Course-In-Mathematical-Analysis-J-C-Burkill-H-Burkill-48-320.jpg)
![3.1] LIMITS 35
Za Let (X, Pp), (Y, ©) be metric spaces. Let f be a function with domain E < X
and with range in Y, and suppose that x, is a limit point of E. Show that, if
lim f(x,) exists for every sequence (x,) in E—{x9} such that x, > Xp, then
n>o
lim f(x) exists.
22,
3. Let (X, p), (Y, 0) be metric spaces, and let E be a subset of X, x, a limit point
of E, yo a point of Y.
Show that, if the function f: E> Y is such that f(x) > yo
as x > Xo, then yp € f(E); and that if, in addition, fis injective, then y, is a limit
point of f(£).
4. Let (X, p), (¥, o), (W,7) be metric spaces and let D, E be subsets of X, Y
respectively. Leth = go f: D > Wbe the composition of the functions f:D > Y,
where f(D) © E, and g: E> W. Also (i) Xp is a limit point of D and f(x) > yo
as x > X9; (ii) yo is a limit point of
E and g(y) > wo as y > yo. Show that, whenf
is injective (so that, by 3, yo is automatically a limit point of E), then h(x) > wo
as x > x9. Construct an example in which A(x) +> wo.
5. Prove that the statements (3.11)-(3.13) hold under the conditions given at the
end of the section.
3.2. Continuous functions
The definition of continuity which we shall now give is a natural
generalization of the concept familiar in elementary analysis.
>
Definition. Let (X, p) and (Y, c) be metric spaces. Then the function
f: X > Yis said to be continuous at the point xy of X if, given e > 0,
there is a 6 > O such that
o(f(x),f(%)) < € whenever p(x, Xo) < 6
(or f(B@%o; 6) = BFC); €))-
If f is continuous at all points of a set, it is said to be continuous on.
that set.
The definition implies that f is continuous at a limit point x) of
xt f) +f) _as_ x >;
and that fis continuous at all isolated points of X (for if x, is an
isolated point and 6 is sufficiently small, the only point x such that
p(X, Xo) < Fis Xp itself).
~Let f be a function
whose domain is a subset X, of a metric space
(X, p). When considering the limits of f we are interested in the limit
points of X, and these need not belong to Xj. However, for questions Pe
<i}
of
continuity,
only the points of the domain X, of f are relevant. re
Since X,, endowed with the metric induced by (YX, p), is itself a
metric space, it is therefore simplest in this context to treat X, as the
2-2](https://image.slidesharecdn.com/22284738-250620074959-26290e84/85/A-Second-Course-In-Mathematical-Analysis-J-C-Burkill-H-Burkill-49-320.jpg)
![36 CONTINUOUS FUNCTIONS ON METRIC SPACES 3.2
fundamental space. This is, in fact, what we did in the definition of
continuity.
Mlustrations. The function f of illustration (i)(c) on p. 32 whose domain is de
set of rational numbers, is discontinuous everywhere. If f,, with domain R* is
ety f(x) when x is rational,
(x) = cone
A 0 when x is irrational,
then f; is continuous at all irrational points and discontinuous at all rational
points. : : . }
The function h: C[0, 1] > C[0, 1] of illustration (iii) on p. 33 is continuous
on C[0, 1]. This follows from the inequality
PCACA,), hs) < PCr» $2)
which holds for all ¢,, 6. of C[0, 1]. :
Let (X, p) be any metric space, let a be a fixed point of X and let the function
g: X > R' be defined by the equation
a(x) = pCa, x).
Since P(@, x1) —P(@, x2) < P(%2, X)
and P(@, X2)— P(A, X31) < PC; Xa),
le@—g@)| = |p, x)—P(@, x2)| < PO, x2)
for all x,, x, in X. Hence g is continuous on X.
Algebraic operations on continuous functions with values in a
normed vector space again yield continuous functions. If the functions
f,g: X > V and the functions ¢, y: X > R! (or Z) are continuous
at xo, then ¢f+yeg is continuous at x; so is f.g if V is an inner
product space, and //gis continuous at x» if Vis R! or Z and g(xp) + 0.
All this is obvious when xy is an isolated point of ¥ and otherwise
follows from the corresponding results on limits. In the next theorem
we return to functions with arbitrary ranges. The result contrasts
with exercise 3(a), 4.
Theorem 3.21. Let (X,p), (Y,o), (W,7) be metric spaces and let
h = gof: X > W be the composition of the functions f: X > Y and
g: Y> W. If fis continuous at x» and g is continuous at yy) = f(x),
then h is continuous at x».
Proof. Given e, there is an 7 such that
7(g(y); 8(Yo)) < € whenever o(y, Yo) < 7.
There is now a 6 such that
o(f(x), f(%)) < 7 whenever p(x, x) < 6](https://image.slidesharecdn.com/22284738-250620074959-26290e84/85/A-Second-Course-In-Mathematical-Analysis-J-C-Burkill-H-Burkill-50-320.jpg)
![3.2] CONTINUOUS FUNCTIONS Kf
and so @ is also such that
(A(x), h(X)) < € whenever p(x, X9) < 6. |
It is important to note that the inverse function of a bijective
continuous function is generally not continuous. A simple example
‘is thefunction with domain Y = [0, 1) u [2, 3)and range Y = [0, 2]
defined by
0) [x for0 <x <1,
x) =
x= for 2x < 3)
If Xand Y have the metrics induced by R!, then fis continuous on X.
However f1, which is given by
y for0 < y <1,
TO) = f
(p41 forl < y < 2,
is discontinuous at the point 1. For other examples see exercises
3(b), 4 and 5.
The concept of continuity can be formulated in a variety of ways.
We shall next consider equivalent definitions, first for continuity at
a point and then for continuity on the domain of the function.
Theorem 3.22. Let (X,p) and (Y,) be metric spaces. A necessary
and sufficient condition for the function f: X > Y to be continuous at
the point Xp is that x, > Xo implies f(Xn) >f(Xo)-
The proof is similar to that of theorem 3.1. The theorem may also
be deduced from theorem 3.1, but slight complications arise, since
members of the sequence (x,) may now be Xo.
Theorem 3.22 shows that the introduction of equivalent metrics
in the domain or the range of a function does not affect the property
“Of
continuity: if p, , are equivalent metrics on X and o, oj are-
equivalent metrics on Y, and if f: X > ¥ is continuous at x» with
Tespect to p and
o, then fis also continuous at xo with respect fo p;
and 0}.
—_
Theorem 3.23. Let (X,p) and (Y,) be metric spaces. Each of the
following conditions is necessary and sufficient for the function
f: X > Y to be continuous on X:
(i) Whenever G is open in Y, then f-(G) is open in X.
(ii) Whenever F is closed in Y, then f(F) is closed in X.
Proof. (i) First suppose that f is continuous on X and that the set
G is open in Y.](https://image.slidesharecdn.com/22284738-250620074959-26290e84/85/A-Second-Course-In-Mathematical-Analysis-J-C-Burkill-H-Burkill-51-320.jpg)
![¢ONTINUOUS FUNCTIONS ON METRIC SPACES B.2
If the set f-1(G) is empty then it is also open. If f(G) is not empty,
let x) be any one of its points. Then f(x) € G and, since G is open,
there is an ¢ such that
yeG whenever o(y,f(%)) < &
But, since f is continuous at Xo, there is a 6 such that
o(f(x),f(%)) < € whenever p(x, Xo) < 6.
Hence f(x)€G whenever p(x, x9) < 4,
ie. xef-(G) whenever p(x, Xo) < 6.
This shows that x) is an interior point of f(G). It follows that
f-(G) is open in X.
Next suppose that, whenever G is open in Y, then f(G) is open inX.
Let x) be any point of X and take any positive e. The open ball
B(f(%); €) is open in Y and so, by hypothesis, the set f{B( f(x); ©)}
is open in X. This set also contains x, and therefore there isa d > 0
such that ef-{B(f(%); )} whenever p(x, Xo) < 6,
ne F(x) € BU); ©) whenever p(x, Xo) < 4,
Le, o(f(x), f(%o)) < € whenever p(x, X) < 6.
Thus fis continuous at x) and so on X.
(ii) This part follows from (i) by the method of complements since,
if E is any subset of Y,
XE) nfE)= @ and f(E)u fe) =X. |
In the last theorem the inverse images may not be replaced
by direct
images. It is not true that /(S) is open (closed) in Y whenever Sis open
(closed) in ¥. The function defined on p. 37 will illustrate this
(exercise 3(b), 6) but it is instructive to employ also more common-
place examples. For instance the function f with domain R! and
range R! given by AO = Hee
av maps the open interval (0, 1) onto (0, 3]. Next, let g, with domain
(— ©, 00) and range [—4, 4] be defined by
x
ex
g(x) =
This function maps the closed interval [1, 00) on the interval (0, 4]
which is not closed in [—4, 4].](https://image.slidesharecdn.com/22284738-250620074959-26290e84/85/A-Second-Course-In-Mathematical-Analysis-J-C-Burkill-H-Burkill-52-320.jpg)
![3.2] CONTINUOUS FUNCTIONS 39
We have previously mentioned that a bijective continuous function
F: (X, p)> (Y, ¢) need not be such that f: (Y, c) > (X, p) is con-
“tinuous. When
/— is continuous, fis called a homeomorphism. (Note
that then f~ is also a homeomorphism.) Two metric spaces are said
to be homeomorphic if there exists a homeomorphism on one into
the other. If f:(X,p)—>(Y,¢) is a homeomorphism, then, by
theorem 3.23, f maps open (closed) sets in X into open (closed) sets
in Y and there is a bijection between the open (respectively closed)
sets in the two spaces.
A bijective function f: (X, p) > (Y,
2) such that
oF)» f&%2)) = PP 2)
for all x;, x, € X is called an isometry. It is plain that an isometry
is a homeomorphism. Two metric spaces are called isometric if there
exists an isometry on one into the other. Such spaces have identical
metric properties although their elements may be of entirely different
kinds. For instance R} is isometric with the subspace A of C[0, 1]
consisting of the functions ¢, (— «0 < A < ) defined by
d(x) = Ax O<x< 1).
Another example of an isometric pair is provided by R®, Z.
Linear functions. We briefly consider a class of functions which
plays an important part in modern analysis.
Definition. Let V, W be two real (complex) normed vector spaces. A
function f: V > W is called linear if
S (GX, + HX) = Of (Xr) + of (2)
for all x, X, € V and all real (complex) numbers 04, %».
Clearly f(@) = 9, where the letter 0 on the left stands for the zero
element in V and that on the right for the zero element in W. (It is
not customary to use distinctive notations for the zero element and
the norm in the two spaces.)
Mlustrations
(1) The function f: C[a, b] > R} defined by the relation
£) = fo dt (pe Cla, b)
is linear. :
(2) The reader is probably acquainted with linear functions on one Euclidean
space into another. We shall use three properties of these functions which are](https://image.slidesharecdn.com/22284738-250620074959-26290e84/85/A-Second-Course-In-Mathematical-Analysis-J-C-Burkill-H-Burkill-53-320.jpg)
![40 CONTINUOUS FUNCTIONS ON METRIC SPACES [3.2
proved in most books on linear algebra. (See, for instance, L. Mirsky, Intro-
duction to Linear Algebra, chapters IV and V.)
(i) The function f: R" + R” is linear if and only if it is of the form
LG, 0103 Xm) = Vas v9 Yn)r
where V=au%it. nent 21)
Vn = An Xt...
+ AnmXm
Va Qy1---Bym [%1
or Noa Sipe v
a ONE Cell Vee
The nx m matrix (a,;) is said to represent f.
(ii) If the linear functions f: R” > R” and g: R" > R® have matrices A, B,
then the linear function go f: R” > R? is represented by the matrix BA.
ii) Let f: R"™ > R* be a linear function with matrix A. Then f is injective if
and only ifm > mand A has (maximum) rank m. Also fis surjective (i.e. the range
of fis the whole of R”) if and only if < mand A has (maximum) rank n. Thus,
a necessary and sufficient condition for f to be bijective (so that f— exists and is
again a linear function) is that m = n and A has maximum rank, i.e. det A + 0.
When f is bijective, f-! has matrix A“.
It is easy to see that a linear function on R” into R” is continuous
at all points of its domain (see exercise 3(b), 8). A consequence of
the theorem below is that any linear function either is everywhere
continuous or is nowhere continuous.
Ne Theorem 3.24. Let V, W be normed vector spaces. If the function
y JS: V + Wis linear, then the following three statements are equivalent.
Shi (i) fis continuous on V.
ire) | (ii) There is a point x») € V at which f is continuous.
(iii) ||f(%)||/||
|| is bounded for x € V—{6}.
pey 5
ja A a Proof. (ii) = (i). Take any ¢ > 0. Since f is continuous at xp,
there is a 6 > 0 such that ||x— xl < 6 implies || f(x) —f(x)|| < «.
Let c be any point of V and take any x such that ||x—cl| < 6. Then
|(@-e+x)—xl| < 6 and therefore ||f(x-—c+x)—f(x)| < 6 ie.
|f0C)—f(0)|| < €. Thus fis continuous at c.
(i)+ (iii). Suppose that ||f(x)||/||x| is not bounded. Then, given
any integer 7, there is a point x, € V—{6} such that ||f(x,,)]|/||xnl| > 7.
If v, = (||x,||)-1x,, then ||v,|| = 1/n and
_ If@nl,
|Fe,)ll “nx >
Hence fis not continuous at 0.](https://image.slidesharecdn.com/22284738-250620074959-26290e84/85/A-Second-Course-In-Mathematical-Analysis-J-C-Burkill-H-Burkill-54-320.jpg)
A Second Course In Mathematical Analysis J C Burkill H Burkill
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- 5. A Second Course in Mathematical Analysis J.C. Burkill and H. Burkill — | Cambridge Mathematical Library
- 7. A Second Course in Mathematical Analysis
- 8. Other books available in the Cambridge Mathematical Library G.E. Andrews A. Baker H.F. Baker RS. Ball N. Biggs S. Chapman & T.G. Cowling R. Dedekind K. Falconer & C.A. Rogers G.H. Hardy G.H. Hardy, J.E. Littlewood & G. Pélya D. Hilbert W.V.D. Hodge & D. Pedoe R.W.H.T. Hudson A.E. Ingham B. Jeffreys & H. Jeffreys H. Lamb J. Lighthill M. Lothaire F.S. Macaulay L.C.G. Rogers & D. Williams W.T. Tutte G.N. Watson A.N. Whitehead & B. Russell E.T. Whittaker E.T. Whittaker & G.N. Watson A. Zygmund The theory of partitions Transcendental number theory Abelian functions A treatise on the theory of screws Algebraic graph theory, 2nd edition The mathematical theory of non-uniform gases Theory of algebraic integers Hausdorff measures A course of pure mathematics, 10th edition Inequalities, 2nd edition Theory of algebraic invariants Methods of algebraic geometry, volumes I, II & III Kummer’s quartic surface The distribution of prime numbers Methods of mathematical physics Hydrodynamics Waves in Fluids Combinatorics on words The algebraic theory of modular systems Diffusions, Markov processes and Martingales volumes I & IT Graph theory A treatise on the theory of Bessel functions, 2nd edition Principia Mathematica to *56 A treatise on the analytical dynamics of particles and rigid bodies A course of modern analysis, 4th edition Trigonometric series
- 9. A SECOND COURSE IN MATHEMATICAL ANALYSIS BY J-C.BURKILLI, F.R:S: Master of Peterhouse, Cambridge AND H. BURKILL Senior Lecturer University of Sheffield 1 S = 4 ese aye CAMBRIDGE AT THE UNIVERSITY PRESS 1970
- 10. PUBLISHED BY THE PRESS SYNDICATE OF THE UNIVERSITY OF CAMBRIDGE The Pitt Building, Trumpington Street, Cambridge, United Kingdom CAMBRIDGE UNIVERSITY PRESS The Edinburgh Building, Cambridge CB2 2RU, UK 40 West 20th Street, New York, NY 10011-4211, USA 477 Williamstown Road, Port Melbourne, vic 3207, Australia Ruiz de Alarc6n 13, 28014 Madrid, Spain Dock House, The Waterfront, Cape Town 8001, South Africa http://www.cambridge.org © Cambridge University Press 1970 This book is copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published in the Cambridge Mathematical Library 2002 Printed in the United Kingdom at Atheneum Press Ltd, Gateshead, Tyne & Wear A catalogue record for this book is available from the British Library Library of Congress Cataloguing in Publication Data ISBN 0 521 52343 5 paperback
- 11. CONTENTS Preface page vii 1. 2. 4. 16 SETS AND FUNCTIONS il 1.1 Sets and numbers, p. 1. 1.2. Ordered pairs and Cartesian products, p. 5. 1.3. Functions, p. 6. 1.4. Similarity of sets, p. 10. Notes, p. 15. METRIC SPACES 17 2.1. Metrics, p.17. 2.2. Norms, p. 21. 2.3. Open and closed sets, p.25. Notes, p.31. CONTINUOUS FUNCTIONS ON METRIC SPACES 32 3.1. Limits, p. 32. 3.2. Continuous functions, p. 35. 3.3. Connected metric spaces, p. 43. 3.4. Complete metric spaces, p. 47. 3.5. Completion of metric spaces, p. 57. 3.6. Compact metric spaces, p. 60. 3.7. The Heine—Borel theorem, p. 67. Notes, p. 70. LIMITS IN THE SPACES R! AND Z © 4.1, The symbols O, 0, ~, p.75. 4.2. Upper and lower limits, p.77. 4.3. Series of complex terms, p. 84. 4.4. Series of positive terms, p. 86. 4.5. Conditionally con- vergent real series, p. 89. 4.6. Power series, p. 93. 4.7. Double and repeated limits, p.95. Notes, p. 100. UNIFORM CONVERGENCE 103 5.1. Pointwise and uniform convergence, p.103. 5.2. Properties assured by uniform convergence, p. 109. 5.3. Criteria for uniform convergence, p. 116. 5.4. Further properties of power series, p. 122. 5.5. Two constructions of continuous functions, p. 127. 5.6. Weierstrass’s approximation theorem and its generalization, p. 132. Notes, p. 137. INTEGRATION 139 6.1, The Riemann-Stieltjes integral, p. 139. 6.2. Further properties of the Riemann— Stieltjes integral, p. 148. 6.3. Improper Riemann-Stieltjes integrals, p. 156. 6.4. Functions of bounded variation, p. 162. 6.5. Integrators of bounded variation, p. 168. 6.6. The Riesz representation theorem, p. 173. 6.7. The Riemann integral, p.178. 6.8. Content, p. 181. 6.9. Some manipulative theorems, p. 187. Notes, p. 195. FUNCTIONS FROM R™ TO R” 197 7.1. Differentiation, p. 197. 7.2. Operations on differentiable functions, p. 204. 7.3. Some properties of differentiable functions, p.208. 7.4. The implicit function theorem, p.212. 7.5. The inverse function theorem, p. 222. 7.6. Functional depend- ence, p. 226. 7.7. Maxima and minima, p. 231. 7.8. Second and higher derivatives, p. 237. Notes, p. 244,
- 12. vi 10. 11. 1b 1: 14. CONTENTS INTEGRALS IN R” page 246 8.1. Curves, p.246. 8.2. Line integrals, p. 252. 8.3. Integration over intervals in R", p. 258. 8.4. Integration over arbitrary bounded sets in R", p. 265. 8.5. Differentia- tion and integration, p.274. 8.6. Transformation of integrals, p.278. 8.7. Functions defined by integrals, p. 289. Notes, p. 297. FOURIER SERIES 299 9.1. Trigonometric series, p. 299. 9.2. Some special series, p. 302. 9.3. Theorems of Riemann. Dirichlet’s integral, p. 307. 9.4. Convergence of Fourier series, p. 311. 9.5. Divergence of Fourier series, p. 315. 9.6. Cesaro and Abel summability of series, p.320. 9.7. Summability of Fourier series,p. 323. 9.8. Mean square approxi- mation. Parseval’s theorem, p.327. 9.9. Fourier integrals, p. 330. Notes, p. 335. CoMPLEX FUNCTION THEORY 338 10.1, Complex numbers and functions, p. 338. 10.2. Regular functions, p. 341. 10.3. Conformal mapping, p.345. 10.4. The bilinear mapping. The extended plane, p. 346. 10.5. Properties of bilinear mappings, p. 348. 10.6. Exponential and logarithm, p. 352. Notes, p. 355. CoMPLEX INTEGRALS. CAUCHY’S THEOREM 358 11.1. Complex integrals, p. 358. 11.2. Dependence of the integral on the path, p. 360. 11.3. Primitives and local primitives, p. 361. 11.4. Cauchy’s theorem for a rectangle, p.364. 11.5. Cauchy’s theorem for circuits in a disc, p.367. 11.6. Homo- topy. The general Cauchy theorem, p. 368. 11.7. The index of acircuit for a point, p.370. 11.8. Cauchy’s integral formula, p. 372. 11.9. Successive derivatives of a regular function, p. 375. Notes, p. 378. EXPANSIONS. SINGULARITIES. RESIDUES 381 12.1. Taylor’s series. Uniqueness of regular functions, p. 381. 12.2. Inequalities for coefficients. Liouville’s theorem, p. 383. 12.3. Laurent’s series, p. 386, 12.4. Singu- larities, p. 389. 12.5. Residues, p. 390. 12.6. Counting zeros and poles, p. 392. 12.7. The value z = ©, p. 396. 12.8. Behaviour near singularities, p. 397. Notes, p. 399. GENERAL THEOREMS. ANALYTIC FUNCTIONS 400 1350) Regular functions represented by series or integrals, p. 400. 13.2. Local mappings, p. 404. 13.3. The Weierstrass approach. Analytic continuation, p. 408. 13.4. Analytic functions, p. 411. Notes, p. 414. APPLICATIONS TO SPECIAL FUNCTIONS 416 14.1. Evaluation of real integrals by residues, p. 416. 14.2. Summation of series by residues, p.424, 14.3. Partial fractions of cot z, p. 425.‘ 14.4. Infinite products, p. 428. 14.5. The factor theorem of Weierstrass. The sine product, p. 432. 14.6. The gamma function, p. 436. 14.7. Integrals expressed in gamma functions, p. 440. 14.8. Asymptotic formulae, p. 444. Notes, p. 451. Solutions of Exercises 453 References 522) Index 523
- 13. PREFACE This course of analysis is intended for mathematical specialists in their second and third years at Universities. We assume familiarity with the concept of a limit and its applications to infinite series and to the differential and integral calculus. The contents of A First Course in Mathematical Analysis by one of us (J.C.B.), which do not include Cauchy sequences, upper and lower limits or uniform con- vergence, form a suitable foundation for this book. From time to time we shall need to refer to a basic work on analysis and for simplicity we shall refer to the First Course, shortening its title tol; An undergraduate after his first year should be ready for a more abstract setting and prepared to think in metric spaces instead of the Euclidean line or plane. The study of metric spaces provides not only a means of unifying different topics in analysis but also a natural link with topology. Chapters 1-9 concentrate on general analysis and real functions and 10-14 on complex functions. These last five chapters are largely independent of 6-9, and the reader who wishes to reach Cauchy’s theorem quickly needs little more than §8.1 and §8.2. After careful thought we decided to treat the Riemann and Riemann-Stieltjes integrals fully (chapters 6 and 8) and to leave the Lebesgue integral out. For many purposes the Riemann integral is sufficient, and the inclusion of an adequate account of Lebesgue measure and integration would impair the balance of the book. We are very grateful to Dr L. Mirsky and to Professor G. E. H. Reuter, each of whom has read the entire manuscript. The final form of it owes a great deal to their care and vigilance and to the experience which they gained from their own teaching. J.C.B. August 1968 H.B.
- 14. ae ae
- 15. 1 SETS AND FUNCTIONS 1.1. Sets and numbers It has been recognized since the latter part of the nineteenth century that the idea of number (real and complex), and therefore all analysis, is based on the theory of sets. In modern analysis the dependence is explicit, for the language and algebra of sets are in constant use. The reader is likely to be familiar with the intuitive notion of a set and with the basic operations on sets. In this section we therefore confine ourselves to fixing the terminology and recapitulating the tesults that will be used subsequently. The notes at the end of the chapter refer to books which develop set theory systematically from explicitly stated axioms. There are synonyms for the word set such as collection and space. The members of a set are also called its elements or points. The statement that a belongs to (or is a member of) the set A is written aeA, If a does not belong to A we write aéA. We denote by @ the empty set, namely the set which has no members. Inclusion of sets. Suppose that every member of the set A also belongs to the set B, i.e. that xeEA>xeB. (ari) Then we say that A is contained in B and write A‘< Bor) B >A; the set A is said to be a subset of B. Note that any set is a subset of itself. Also the empty set is a subset of every set; for (1.11) is logically equivalent to Renee and, when A is the emipty set, this implication clearly holds for any set B. 1 Bsc
- 16. #4 SETS AND FUNCTIONS fii If 4 ¢ Band Bc A, ie. xeAoxeB, then A and B have the same members and we write AB. If A < Band A + B, we can call A a proper subset of B. Union and intersection. Given any two sets A, B, the set which consists of all the elements belonging to 4 or to B (or to both) is called the union of A-and B and is denoted by AUB. The set consisting of the elements which belong to both A and B is called the intersection of A and B and is denoted by AnB. The sets A, B are said to be disjoint if they have no elements in com- mon, ie. if ANB= o. Theorem 1.11. The following identities hold for any sets A, B, C. @) AUB= BUA, ANB=BnA4; Gi) AU(BUC) = (AUB)UC, An(BnNO) = (ANB)nC; Gil) AU(BNC) = (AUB)n(AUQ), An(BuUC) = (AnB)u(AnO). Proof. We illustrate the argument used for establishing these identities by proving the first result in (iii). Let x be any element of 4U(BnC). Then xe A or xE BNC. If xe A, then xe AUB and xe AUC, so that xe(AUB)N(AUC). If xe Bn C, then x € Band x € Cso that again xe (AU B) nN (AUC). We have therefore shown that AU(BNC) < (AUB)n(AvO). (1.12) Now let ye(AUB)n(AUC). Then ye AUB and ye AUC. It follows that y < A or that ye Band ye C,ie. that ye Aorye BNC. Hence y € A U(BN C) and so (AUB)n(4uUC) < Au(Bn©). (113) The inclusion relations (1.12) and (1.13) now yield the required identity. |
- 17. 1.1] SETS AND NUMBERS 3 The laws of operation with U and fn on sets have some likeness to those with + and x on numbers. In fact U and n have replaced the signs for sum and product formerly used in the algebra of sets. The likeness is only partial: the identity 4 UA = A has no analogue in numbers; the second law in theorem 1.11 (iii) has an analogue, but not the first. The definitions of union and intersection previously given may be extended. If @ is an arbitrary collection of sets, the union of these sets, denoted by UA Act is the set consisting of all those elements which belong to at least one of the sets A. The intersection of the sets of @, denoted by MN A, Ace is the set consisting of the elements which belong to all the sets 4. We require one more operation with sets. If A, B are any two sets, the difference YER is the set consisting of those elements which belong to A but not to B. Note that the definition does not require B to be a subset of A. How- ever a particularly important case occurs when all sets under con- sideration are subsets of a given set X. Then the set X—A is called the complement of A (relative to X) and we shall denote it by A’. Clearly (A’)’ = A. Theorem 1.12. For any collection @ of subsets of a set X, COPA] fe Sands Cay) =U 42 Ac€ Ac€ Ac€ Ac€ The proof is left to the reader. Theorem 1.12 shows that the operations U and () are comple- mentary and that an algebraic identity involving U and (-) will admit of a dual identity obtained by interchanging U, (). For example, in each pair of identities in theorem 1.11 the second follows from the first. We shall take for granted that the reader is familiar with the systems of real and complex numbers. There are well known and simple methods for obtaining the complex numbers from the real numbers. For our purposes it is only necessary to postulate the existence of the system of real numbers as one satisfying certain axioms which are 1-2
- 18. 4 SETS AND FUNCTIONS [1.1 given in C1 (11-12) and are restated in the notes at the end of the chapter. However, it is possible to construct the real numbers from much more primitive objects. Again the notes contain some remarks on the subject. The set of real numbers will be denoted by R?, the set of complex numbers by Z. Our notation for the open interval in R1 which consists of the points x such that a < x < b is (a, b). The closed interval of points x such that a < x < b is written [a, 5]. The intervals defined bya < x < banda < x < bare written [a, b) and (a, b]. The infinite interval [a, 00) consists of the points x such that x > a. Correspond- ing interpretations are put on the expressions (a, 00), (—©, a), (©, a]. Finally, (— 00, 0) is R. Exercises 1(a) 1. Prove that @ AUB=A=+A>B; Gi) ANB=A+ACB. 2. Prove that (Bu C)n(Cu A)n (AUB) = (BN O)u(CN A)U (AN B). 3. Show that A—B = Af B’. Hence or otherwise prove that @) (A—B)n(A—-C) = A-(BUC) = (A-B)-C; Gi) (A—B)U(A—C) = A-(B0Q); Gii) (A—C)n (B—C) = (An B)-C; (iv) (A—C) U(B-C) = (AU B)-C. 4. Prove that (@) (€A-B)nC= (AnC)-(BnO; (i) (A-B)UC = (AUC)-(BUQ)=Ce= oa. 5. Prove that @ (AUB)-B=A+ANB= Q; (ii) (A-B)UB=A+A>B. 6. The symmetric difference A A Bof the sets A, Bis d ¢ ) ’ N fined as the set of el in A or in B, but not in both. Show that ae 44 B= (A~B)U (B—A) = (AU B)—(4nB) = (AU B)n(4'UB). Deduce that @MASB=9+A=B; (ii) A’ AB=AAB, 7. Prove that @AAB=BAA; (ii) AABAC=AA(BAO,.
- 19. 1.1] SETS AND NUMBERS 5 8. Prove that @ AA BNC= (ANC) A (BNO); @) AA BUC=(AUQABUO=$CeH= oa; ii) (4 A B)U(AA C) = (AU BUC) n(4’U BUC). 9. Show that AAB=CAD#AAC=BAD. 10. When Eis a finite set (i.e. one with finitely many elements), denote by |E| the number of its elements. Show that, if the sets A, B are finite, then |A|+|B] = |4u Bl +]4nB}. 1.2. Ordered pairs and Cartesian products A set with elements a, b, c, ... is often denoted by the symbol only ae (1.21) The notation calls for a number of comments. (i) In such a listing of elements it is immaterial whether a particular element appears once or several times. This convention is purely a matter of convenience. For instance it allows us to denote the set of roots of a complex quadratic equation by {a, b}, whether a + b or a=b. (ii) The order in which a, b, c, ... are written in {a, b, c, ...} has no significance. For example {a, b} = {b, a}. (iii) It is important to distinguish between the object a and the set {a}, i.e. the set whose only member is a. Thus the set {@} has one element, while @ has none. The notation (1.21) has a useful variant. Given a set X, denote by P(x) a statement, which is either true or false, about the element x of X. The set of all x € X for which P(x) is true is written {x € X|P(x)}, or simply {x|P(x)} if the identity of the set X is clear from the context. Illustrations @ {xEZ|x?+1 = 0} = fi, —i; (i) (ce Rx2+1 = 0} = 2; ii) {xe R'la < x < 5} = [a, A). The notion of a set does not involve any ordering among the elements. For instance in the set {a, b}, a and b have the same status.
- 20. 6 SETS AND FUNCTIONS [1.2 The ordered pair (a, b) is the set {a, b} together with the ordering ‘first a, then b’. Thus (a, ) and (6, a) are different unless a = b. (The context will determine whether the expression (a, 5) stands for an ordered pair or an open interval.) ; The intuitive concept of an ordered pair (a, b) may be formalized by the definition (ab) = tah, (a, BN, (See exercise 1(5), 1.) An ordered set (a, ..., @,) of any finite number n (> 1) of elements is defined in a similar way. Definition. Given the non-empty sets X4,...,Xn, their Cartesian LOS, yoy (1.22) is the collection of all ordered sets (x1, ...) Xn) such that Xy CAG, 55 Xp, SC Aq IfX, =... = X, = X, the set (1.22) is denoted by X"; X1is taken to be X. For n > 1, R® = R'x...x R! (n factors) is the set of all ordered sets (x1, ..., X,) of real numbers. We define an interval in R” as a set Ty Xsan MIs where J,, ..., J, are intervals in R'. For example [a, b] x[c, d] is the set of points (x, y) € R? such thata <x <bandc<y<d. 1.3. Functions Let X, Y be two non-empty sets. A relation from X to Y is a subset of Xx Y. Ifa relation f, i.e. a subset of Xx Y, is such that, for every x € X, there is one and only one member (x, y) of f, then fis said to be a function on X to (or into) Y; we express this symbolically by writing Six = YF. The set X is called the domain of f. In some contexts the terms mapping, transformation, or operator are often used as synonyms for function. Let (x, y) be an element of the function f: X > Y. Then we say that y is the value of f at x (or the image of x under f) and we write y = f(x). If Lis any subset of X, the subset of Y (ye Yly =f) and xe} is called the image ofE under /; it is denoted by f(E). The set f(X) is called the range of f, Note that f(X) may be a proper subset of Y.
- 21. 1.3] FUNCTIONS 7 If a function fhas its domain in X and its range in Y, we say that Sis a function from X to Y. A function from R! to R! will be called a real function, a function from Z to Z a complex function. Illustrations (i) We obtain a function on X to Y whenever we have a rule for assigning a single value y in Y to each x of X. Thus the equation x?+y? = 1 defines a function on R! to R! whose range is (— 09, 1]. Gi) The equation x*+y? = 1 defines a relation but not a function on [—1, 1] to [—1, 1] because, whenever — 1 < x < 1, it is satisfied by two values of y. Gii) The logarithmic function is the function on (0, 00) to R' which assigns the value log x to the positive number x. (iv) The sequence ay, a2, as, ... is a function whose domain is the set of positive integers exhibited as suffixes. More generally, given two arbitrary non-empty sets A, Janda function on Jto A, we sometimes find it convenient to write the image of a member / of J in the form a;,. (v) Denote by © the set of all real functions ¢ defined by an equation of the form G@) = axt+fh (xe R), where a, f are real numbers. A function f: R* > ® is obtained by assigning to each point (a, b) € R* the function ¢,,, given by ¢.,(x) = ax+b (xe R%). Let f be a function on X into Y. If f(X) = Y, ie. if every ye Yis the image under fof at least one x € X, then we say that fis surjective, or that fmaps X onto Y. If different elements of X have different images, i.e. if f(x) = f(x’) implies x = x’, then fis called injective. If f is both surjective and injective, so that every value y in Y is assumed once and once only, then fis said to be bijective. A bijective function is also called a bijection; it effects a one-to-one correspondence between its domain and range. Consider again the illustrations above. (i) The function maps R! onto (—, 1]; it is not injective. (iii) The logarithmic function is a bijection on (0, <0) to R*. (v) This function is a bijection on R* to ®. Inverse functions. Suppose that the function f has domain X and range Y. Tof, which is a subset of X x Y, there corresponds the sub- f Yx X defined Sine GNIGIER: (31) Plainly this is a function with domain Y and range X if and only if f is bijective. When f is bijective, (1.31) is called the inverse Function of f and is denoted by f—. For instance the logarithmic function has
- 22. 8 SETS AND FUNCTIONS [1.3 an inverse with domain (—00, 00) and range (0,00); the inverse is the exponential function. If the function f: X > Y is injective, then f: X¥ > f(X) is bijective and f1:f(X)—> X is a function from Y to X. For example the function f: R! > R! defined by ; f) =e has the inverse 1: (0, 1) > R? given by 1 “4(y) = tog (5-1). f70) = log 5 It is clear that, if f has an inverse, then f-1 has an inverse and (yt =F. Inverse images. Let f: X > Y be an arbitrary function and let E be any subset of Y. We denote by f(£) the set {xe X| f(x) € E}. This set is called the inverse image of E. It is important to note that the definition of f“(£) does not presuppose the existence of the inverse function f+. However, when f does exist, then f(£) is the image of £ under f. If f is the function in (i) on p. 7 then, for example, f°, 1) = (-1,)); the function f-! does not exist. Composition of functions. Given the functions f: ¥ > Y and g: Y>W, the function h: X > W defined by h(x) = g{f@)} (we X) is called the composition of g and f and is written go f. For three functions fi: X, > X, fa: X,> Xs, fg:X3 > Xa we clearly have ho(hof) = (hofoh so that the expression f,0 f,0 f, has a meaning. The process of com- position may be extended to any number of functions. For a non-empty set A, denote by i, the identity function on A defined by i(x) =x (xe A). If, now,f: X > Yis any function, foix=f, iyof=f; and, when f is bijective, fof = ix, fof = iy.
- 23. 1.3] FUNCTIONS 9 Theorem 1.31, (i) If the function f: X + Y is surjective and if there is a function &: Y>X such that go f = ix, then f is bijective and g = f>. (ii) If the function f: X + Yis bijective and if the function g: Y > X is such that fo g = iy, theng = f-. Proof. @ IEf@) = f(x), then x = g{f(x)} = aff} = x’ and so fis injective as well as surjective. We now have g = goiy = go(fof) = (gof)oft = ixoft =f. (i) g = ixog = (flof)og = fto(fog) =froiy =f". | In theorem 1.31 (ii) the bijectiveness of f is not a consequence of surjectiveness, as it is in part (i). For instance let X = {a, b} (a + b), Y = {c} and let f, g be defined by f(a) = f(b) = c, g(c) = a. Thenf is surjective and fo g = iy, but fis not bijective. If the functions f: ¥ > Y, g: Y > W are bijective, then go/f: X — W is bijective. It follows from theorem 1.31, or it may be proved directly, that (gof)3 = frog. Restriction and extension of functions. Suppose that X, > X, and that f: X, > Y, g: X, > Y are functions such that f(x) = g(x) for all x € X,. We then say that g is the restriction of f to X, and that fis an extension of g to X,. For example let f,: R! > R' and f,: R! > R? PE RTEU HN T A) p fts) — [x] Ae RY, The function g: [0, 00) > R! defined by a(x) =x (x 20) is the restriction to [0, 00) of both f; and f2; and f,, f, are both extensions of g to Ri. Exercises 1(5) 1. Show that {{a}, {a, b}} = {{c}, {c, d}} if and only if a = cand b = d. 2. Prove that @ (AUB)xC= (Ax C)uU (Bx), (ii) (An B)x C = (Ax C)n (Bx), (iii) (A—B)x C = (Ax C)-(Bx C).
- 24. 10 SETS AND FUNCTIONS [1.3 3. Prove that Ax B = BxA if and only if A = B. Is (Ax A)x A necessarily the same set as A x (Ax A)? 4. Show that {a}x {a} = {{{a}}}. 5. Let be a collection of subsets of a set X. Prove that, for any function ff: XY, @fCU A= USA, @ SCN AS JOG Acé Ace Ac€ Show that, if fis injective, then identity holds in (ii); and that fis injective if (AN B) = f(A) nf) for all subsets A, B of X. 6. Show that, if f: X¥ > Y is any function, then f(A-B) > f(A)—f(B) for all subsets A, B of X; and that identity holds for all A, B if and only if fis injective. 7. Let @ be a collection of subsets of a set Y. Show that, for any function f:X7 Y, i OW B= UL), @s> Care) = 0 f7*@. Bee Bee Bee 8. Show that, if f: X¥ > Y is any function, then FXC=D) =f*O=F 1B) for all subsets C, D of Y. 9. Show that, if f: X > Y is any function, then @ F°GU) > 4, Gi) (FB) < B for all subsets A of X and all subsets B of Y. Prove also that in (i) identity holds for all A if and only if fis injective; and in (ii) identity holds for all B if and only if f is surjective. 10. The functions ft: X,; > X2, fo: X.> X3, fg: X3—> X, are such that the compositions f,0 f; : X; > X3 and fo fy: X_, > X, are both bijective. Show that: Sy fa, f are all bijective. 11. Let R be a relation from X to X (i.e. a subset of ¥x X) and write xRy when (x, y) € R. The relation is said to be (i) reflexive if xe X + xRx; Gi) symmetric if xRy > yRx; (iii) transitive if xRy and yRz > xRz. Criticize the following ‘argument’. The relation R is known to be symmetric and transitive. Writing x for z in (iii) and using (ii) we obtain xRy = xRx. Hence R is reflexive. 1.4. Similarity of sets If there is a one-to-one correspondence between two sets A, B, i.e. if there exists a bijection f: A > B, then the two sets are said to be similar and we write Maer.
- 25. 1.4] SIMILARITY OF SETS 11 It is easy to see that similarity has the following three properties: (i) A ~ A (reflexivity); (ii) if A ~ B, then B ~ A (symmetry); (iii) if A ~ Band B ~ C, then A ~ C (transitivity). If we confine ourselves to a given collection @ of sets, then simi- larity is a relation from @ to @ and, in view of (i)-(iii) above, it is an equivalence relation. The reader has probably met such relations in other branches of mathematics, particularly algebra. Finite sets are similar if and only if they have the same number of elements. This implies, in particular, that a finite set cannot be similar to a proper subset of itself. We shall see that infinite sets always contain proper subset: hich they are similar (theorem 1.42, corollary). A simple example is the following. Let A be the set of all integers and let B be the set of all even integers. Then A ~ B, since the function given by y = 2x (xe A, ye B) establishes the required one-to-one correspondence. A set which is similar to the set {1, 2, ...} of positive integers is said ee ee Such a set is therefore one that can be arranged as a sequence. A set is called countable if it is either finite or countably infinite. We now show that there are infinite sets which are not countable. Theorem 1.41. The set of real numbers in the interval (0, 1) is not countable. Proof. Suppose that the real numbers in the interval (0, 1) do form a countable set. This set is certainly infinite and can therefore be arranged as a sequence Ay, Az, Ag, «+++ Now represent each number a, as an infinite decimal An = 0.0n1%ne%ng + in which recurring 9’s are not used. Such a representation is unique. Define h f when @,, + 1, Then b=0.f, hfs -.. is a real number between 0 and 1 which differs from a, in the nth decimal place. Hence b is not one of the numbers @;, ds, ds, ... and this contradicts the original assumption that all real numbers in the interval (0, 1) appear in the sequence (a,). | 2 when«,, = 1.
- 26. 12 SETS AND FUNCTIONS [1.4 Countably infinite sets are the ‘smallest’ infinite sets in the sense of the theorem below. Theorem 1.42. Every infinite set contains a countably infinite subset. Proof. Let A be an infinite set. Take some element of A and call it a,. The set A—{a,} is not empty; denote one of its elements by ap. This process may be continued indefinitely: at the mth stage the set A—{ay, ...) Anat cannot be empty since A is infinite. The set {a,, a2, ...} is a countably infinite subset of A. | Corollary. Every infinite set contains a proper subset to which it is similar. Proof. Let A be an infinite set and let S = {a,, a;, ...} be a count- ably infinite subset of A. The function f: A > A—{a,} defined by An, if %— Gt — lee) fs) = [n+ ( lx ifxeA-S is a bijection. Thus 4 ~ A—{a,}. | The next theorem is frequently used. Theorem 1.43. Every subset of a countable set is countable. Proof. Let A be a countable set and let B be a subset of A. When B is finite there is nothing to prove. Assume therefore that B is infinite, so that A is infinite and, being countable, can be arranged as a sequence (a,,). Since every set of positive integers has a least member, there is an element a,, of B with least suffix n,. Next let a, be the element of B—{a,,} with least suffix n,. The procedure may be repeated indefinitely since B is not finite. The sequence Gages Qn Ogartns (1.41) consists of elements of B and, in fact, exhausts B. For take any element of B. It is of the form a,, since it is a member of A, and it must there- fore occur among the first k elements of the sequence (1.41). Hence B is countable. | Theorem 1.44. The set P? of all ordered pairs (p, q) of positive integers is countable.
- 27. 1.4] SIMILARITY OF SETS 13 Proof. The elements of P%, all of which appear in the array (1, D, G, 2), C, 3). (2, 1), 2; 2), (2, 3), + (3, 1), G, 2), G, 3), -.-, may be arranged as a sequence in several simple ways. Perhaps the simplest is the enumeration by diagonals: G0); 2, (2); GD, 22), 0,3); «3 1), @-1, 2)... Asn); | Exercise. Show that the set P" (n > 2) of all ordered sets (p,, ..., Dn) of n positive integers is countable. pee Theorem 1.45. The union of a countable collection of countable sets is countable. Proof. Let {A, Ao, ...} be a finite or countably infinite collection of countable sets. If =i B= A, B,=An—U Ay (2 > 1), k=1 then the B, are mutually disjoint and UA, =UB, =C, n n say. (The non-committal notation U is used since the number of sets n may be finite or infinite.) By theorem 1.43 each of the sets B,, is countable. Therefore, if B,, is not empty, its elements may be arranged in a possibly terminating sequence Daas Bras «+++ Now let S be the subset of P? which contains a pair (p, q) if and only if B, is either infinite or has at least gmembers. By theorems 1.43 and 1.44, S is countable. Since there is an obvious one-to-one correspon- dence between S and C, it follows that C is countable. | Theorem 1.46. (i) The set of all rational numbers is countable. (ii) The set of all irrational numbers is not countable.
- 28. 14 SETS AND FUNCTIONS [1.4 Proof. A given positive rational number is uniquely representable in the form p/q, where p,q are coprime positive integers. With this number we now associate the ordered pair (p,q). Thus the set of positive rational numbers is similar to a subset of P? and is therefore countable. Part (i) now follows easily by use of theorem 1.45. Since the set of real numbers in (0, 1) is not countable, by theorem 1.43 the set R? of all real numbers cannot be countable. If now the set of irrational numbers were countable, (i) and theorem 1.45 would show R! to be countable. | Instead of considering all rational and irrational numbers we may restrict ourselves to those in any given interval. It follows immediately from theorem 1.43 that the analogue of (i) holds. The analogue of (ii) can then be proved as before by using exercise 2 below. Exercises 1(c) 1. Let C be a countable set. Show that, if A is any infinite set, then AUC~ A; and that, if B is an uncountable set, then B-C~ B. 2. (i) Show that all finite open intervals in R! are similar. (ii) Show that all open intervals in R? are similar. (iii) Show that all intervals in R! are similar. 3. Let @ be a countable collection of disjoint sets such that, for all Ae@, A ~ R’. Show that U_ Aw Ri Ac 4. Show that, if the sets A;,..., A, are countable, then so is 4,x...x A,. (In particular, if Q is the set of rational numbers, then Q” is countable.) 5. Let F be the set of all sequences of 0’s and 1’s. Using the binary representation of real numbers prove that # is similar to the interval (0, 1). Show also that 7? ~ F and deduce that, if the sets 4,, ...,.4, are similar to R', then A,x...x A, ~ R4. (In particular, R" ~ R4.) 6. Prove that the set of all polynomials AyX*+ayxX™ +... +n yX+an with integral coefficients is countable. Deduce that the set of algebraic numbers is countable. (An algebraic number is a number which is a root of an algebraic equation with integral coefficients.) 7. Let R; (0 < i < 9) be the set of those real numbers in (0, 1) which do not use the numeral iin their decimal representation. Prove that R; ~ (0, 1). (Suppose that recurring 9’s are not used.)
- 29. 1.4] NOTES 15 8. A set E of real numbers is such that every series = 2 Xe n=1 whose terms are distinct elements of E, converges. Show that E is countable. 9. Given a set A, denote by “4 the set of subsets of 4 (including @ and A). Show that @ if A has n elements, then 4 has 2” elements; (i) if A is countably infinite, then %4 ~ R}; oe if A is any set, then A + %4 (though clearly A is similar to a subset of “A)» 10. Show that any collection of disjoint open intervals of R' is countable. Deduce that, if J is any interval in R' and the function f: J > R* is monotonic, then the set of points of discontinuity of f is countable. (if ce¢J and c is not an end point, then fet) = lim f(x), f(e—) = lim f(x) z>c+ ze exist; see C1, 32.) NOTES ON CHAPTER 1 §1.1. Axiomatic set theory. The intuitive notion of a set, though perfectly adequate for everyday use by analysts, is, in fact, self-contradictory. To demon- strate this we describe the famous Russell paradox. A set may or may not be a member of itself: most sets are not and we call these normal; an example of a set which is a member of itself is the set consisting of all infinite sets (for clearly there are infinitely many of these). Consider the set N of all normal sets. Is N normal or not? If N is normal, then, by definition of normality, N ¢ N; but, by definition of N, a set which does not belong to N is not normal and we have a contradiction. The supposition that N is not normal is equally untenable. The mathematician’s way out of this dilemma is to construct an axiomatic system designed to mimic the desirable features of the intuitive approach and to avoid its pitfalls. The axiomatic theory of sets is described in Theory of Sets and Transfinite Numbers by B. Rotman and G. T. Kneebone. A more informal atti- tude is adopted by P. R. Halmos in Naive Set Theory. The theory of sets is the starting point of the monumental Eléments de Mathématique, which is designed to present the whole of pure mathematics in strictly logical order. This work, by a group of French mathematicians writing under the collective name of N. Bourbaki, has been appearing in sections since 1939 and is not yet (in 1969) complete. The real number system. The set R’ of real numbers, together with the operations of addition and multiplication and an ordering relation, is an ordered field: it satisfies the following six algebraic and three ordinal axioms. A1. To every pair of real numbers a, b correspond a real number a+, the sum of a and b, and a real number ab, the product of a and b. A2. a+b = b+a; ab = ba. A3. (at+b)+c = a+(b+0); (ab)e = abe). A4. There are real numbers 0, 1 such that O+a=a and la=a for all real numbers a.
- 30. 16 SETS AND FUNCTIONS AS. For every real number a, there is a real number x such that a+x = 0. For every real number a, other than 0, there is a real number y such that ay = 1. A6. (at+b)c = act+be. O1. For every two real numbers a, b, one and only one of a>b, a—b, bea is true. O02. Ifa> bandb>c, thena>c. O3. If a > b, then a+c > b+c; if also c > 0, then ac > be. What distinguishes the real number system from other ordered fields is Dedekind’s axiom: If L, R are two non-empty sets of real numbers such that LU R = R' and every member of L is less than every member of R, then either L has a largest member or R has a least member. The construction of such a richly endowed system is, not surprisingly, a major undertaking. The first step is to obtain the natural numbers 0, 1, 2, ...; and the most satisfying way of doing this is to use only the machinery of set theory (see, for instance, Halmos’s Naive Set Theory). Another possibility is to begin with the set of natural numbers, but to assume no more than a few simple properties (Peano’s axioms). This approach is adopted by L. W. Cohen and G. Ehrlich (The Real Number System), E. Landau (Foundations of Analysis) and H. A. Thurston (The Number System). There are now standard algebraic methods for proceeding first to the integers (0, +1, +2, ...) and then to the rational numbers. At this stage we have arrived at an ordered field which does not, however, satisfy Dedekind’s axiom. There are two principal constructions for the final step to the real numbers. The method of cuts is mentioned in C1 and is given in detail by Landau. Cauchy (or fundamental) sequences are used by Cohen and Ehrlich and by Thurston. This process is also described in the notes at the end of chapter 3, since it is closely related to some of the contents of that chapter.
- 31. 2 METRIC SPACES 2.1. Metrics Classical analysis deals with sets of points in and functions defined on Euclidean space or the complex plane, while in “more recent developments prominent parts are played also by other types of spaces. These spaces lead to theories with many similar features. The reason is that the spaces share the same underlying structure. It is therefore both illuminating and also economical of effort to develop many aspects of analysis in a general setting which includes as special cases the various spaces of particular importance. A fundamental property which these spaces have in common is that in all of them there is a distance between any two points. We now lay down the conditions which such a distance function or metric must satisfy. Definition. Suppose that X is a (non-empty) set and that p is a real valued function on X x X with the following three properties. M1. p(x,y) = O for all x,y eX and p(x,y) = 0 if and only if x=Y3 M2. p(x, y) = p(y, x) for all x, ye X; M3. p(x, z) < p(x, y)+p(, 2) for all x,y,z€X (the triangle inequality). The function p is called a metric or distance on X; and X, taken together with the metric p, is called a metric space which we denote by (X, p). The notation (X,p) emphasizes the fact that the set X and the metric p have equal shares in the construction of the metric space. A given set may give rise to many different metric spaces by having different metrics associated with it. Nevertheless, when it is clear which metric is being used we may, for brevity, write X rather than (X, p) for the metric space. Tllustrations @ Any set X can be equipped with a metric; 0 ifx=y, Bee ifx+y is such a metric, but not a very interesting one. (This metric is called the discrete metric.)
- 32. 18 METRIC SPACES [eeu Gi) On R? the usual metric is given by p(x, ») = |x-yI- The usual metric on R" is defined as follows. If x= Gow, PH One eke px, ») = M@a— yi)? +... +n In). (2.11) The case n = 1 gives the earlier definition of the usual metric on R’. Plainly p satisfies M1 and M2. Putting a; = x:—yi, b; = y:—Z; we see that M3 is equivalent to n t n + (a a+b) < ( > a) +(% 7) if (2.12) = i=1 i= By squaring both sides of (2.12) we obtain yet another equivalent form, namely n n t/n + Sais (3 a) (3 vi) : 2.13) i=1 i=1 i= This now follows from the fact that, for all real &, 3 (a€+b)? = A?+2HE+B > 0, i=1 so that H? < AB, n n n where A= a, H= Yad, B= > i. i=l i=l = Therefore p, defined by (2.11) is a metric. The metric space (R", p), where p is the usual metric, is called n-dimensional Euclidean space. (iii) On Z, the complex plane, the usual metric is of the same form as that on R*. If z = x+iy and z’ = x’+iy’ (x, y, x’, y’ real), PG, 2’) = |z—z'| = Me—-xP+0-y93. In fact, when equipped with their usual metrics, Z and R? are essentially the same metric spaces in which only the notations differ. (iv) Let Bla, 6] be the set of bounded real functions on the interval [a, b] and, for f, Bia, b), | Thee Bahl ope) = sup |f)—ea)- asr<bh Clearly p has the properties M1 and M2. M3 follows from the inequalities sup |A(x)+ ¥()| < sup (|$@)|+ |¥@)|) < sup |4(2)|+sup |¥Q)]. (v) Let S be the space of sequences x = (x, Xa, ...) of complex numbers such that Xx, converges absolutely. For x, y € S put Ea PY) = Ds [Xn—Yal- n=1 It is easy to show that p is a metric on S. Let (X, p) be a metric space and let Y be a subset of XY. If o is thicy restriction ofp to Yx Y, ie. o(x, y) = p(x, y) for all x, ye Y, then.
- 33. 2.1] METRICS 19 ¢ is a metric on Y; it is said to be the metric induced by (X, p) on Y. The metric space (Y, 0) is called a (metric) subspace of (X, p). For instance let R? have its usual metric p given by P(x, Y) = V{Ci— yd? + 2 — Ya)? + %s—Ys)"}- If Y is the set of points x € R* with x; = c, the metric o induced on Yis given by — o(x, y) = V{Ca-y)*+ (yo In future, unless there is a statement to the contrary, the metrics on R” and Z will be taken to be the usual metrics; and subsets of R” and Z will be taken to have the metrics induced by the usual ones. Sequences of points. The following definition extends the notion of convergence of sequences of real numbers to sequences of points in a metric space. Definition. The sequence (x,) of points in the metric space (X, p) is said to converge to the point x(€ X) if P(Xn, xX) +0 as n>. (2.14) We write, as in classical analysis, X, > x as n> or lim X, = x. n> Since p(x, y) is a real number, we are able to use the concept of convergence in R!. We could replace (2.14) by the phrase ‘given e > 0, there is an m such that p(x, x) < € for n > m’. This form of the definition reduces to the familiar definition of a convergent sequence of real numbers when (X, p) is R' with its usual metric. Exercise. Prove that 4 sequence cannot converge to two distinct limits. Illustrations y RI = n 2n* ) @ In let Xn = nti? n—2)° Then, as n > 00, Xn => (8, 2); since PGrm 2) = ee aye Seal me Gi) In Bil, 2] with pg) = sup _|f(x)—g(), let f, be given by 1<a2<2 f(x) = (+x (1 < x < 2). Then Sih where GQ) —* =x = 2), For, when 1 < x < 2, 0 < fl) —f 0) = xO? + I)" — x < x(2¥8—1) < 2(2""—-1) and so Pat) < 224"-1)>0 as n>.
- 34. 20 METRIC SPACES [2.1 When the metrics p, 7 on a set X are such that a sequence con- verges in (X, p) if and only if it converges in (X, a), then p and o are said to be equivalent metrics. We shall see, as the theory develops, Tee dane @ oe eonivaeat mates preserves many of the properties of a metric space. A sufficient, though not necessary condition for the metrics p, 7 to be equivalent is the exi: e of strictly positive constants A, sie as Aolx, Y) < o(x, 9) < Hplx, Y) for all x, ye X. (See exercise 2(a), 7.) ie Exercises 2(a) . Is the function p given by p(x, y) = |x?—y| a metric on (@) (—~, ~), (6) [0, 20)? 2. Let C be the set of bounded and continuous real functions on R! and let p be defined by +h [Peo-vor a. = pé, 4) = sup -o<r<0 0<A<1 Show that p is a metric on C. 3. Let X be a non-empty set and let p be a real-valued function on Xx X such that (@) p(x, y) = 0 if and only if x = y; and Gi) p(x, ») < p(x, z)+ 0, z) for all x, y, ze X. Prove that p is a metric on X. 4. In R" let x, = (€, ..-> nx). Show that Xp > x = (E;, «.:, Ea) as k > oo if and only if, for 1 < i < n, &, > &,. 5. In a metric space (X, p), x, > x and yn > y as n > 0. Prove that Pn» Yn) > PC, Y). 6. Show that, if p is a metric on X, then so is o given by p(x, y) CD) aia pce) and that p, o are equivalent metrics. 7. Prove that the metrics p,o on X are equivalent if there are constants A, # > 0 such that Aolx, y) < a(x, y) < Hplx, y) for all x, y ¢ X. Give an example to show that the converse is false.
- 35. 2.1] METRICS 21 8. On R®, o and 7 are defined by a(x, y) = max (|x:—yi], |x2—yal), 7,9) = [x1 — yal +] %2—yal Show that o and 7 are metrics which are equivalent to one another and to the usual metric p. Compare the sets of pointsx in R® given by p(x, a) < 1,0(x, a) < 1,7(x,a) <1 Tespectively. 9. Denote by ¢ the discrete metric on R! (p. 17, illustration (i). Prove that o and the usual metric p on R! are not equivalent. 10. Let C be the set of real functions continuous on the interval [0, 1] and define ene m +) Laparaet phe) = sup [fe)-so, 0<e<1 ‘1 + otha) = ([ 1rco-eta)* dx) . ‘1 rhe)= | 1f)-et0) ax (A function continuous on a closed interval is bounded and integrable.) Prove that o and 7 are metrics on C. Show also that, for all f, g € C, PL, 8) > OC,8) > TU, 8). For n = 1, 2, ..., the functions f,, g, € C are defined by l-nx O<x<7nr), n(i—nx) (0<x <n), (x) = { &nlx) = { ) i el) 0 rs <ix.<, 1); Prove that the sequence (f,,) converges in (C, o) and in (C, 7), but not in (C, p); and that (g,,) converges in (C,7), but not in (C, p) or in (C, ¢). 11. Let (X4, p,), ..., (Xn, Pn) be any metric spaces. Verify that metrics on X,x ... x X, are given by o(x, y) = max {P,(%1, Va), «++» Pn(Xns Yn)}, Ox, ¥) = VPI, Vi) t+... + PA Xn» Ynd}, OX, ¥) = PilXr, V+... +Pnl%ns Yn)s where x = (x1, ..-5 Xn), ¥ = Oy -++5 Yn) and xi, ¥i € X; @ = 1, ..., 2). Show also that 0, 72, 03 are all equivalent. 2.2. Norms The theory of vector spaces is partly algebraic and partly analytical. We need only the most basic algebraic ideas and these are, no doubt, known to the reader. But, largely to fix notation and terminology, we formally define a vector space. Definition. Let V be a non-empty set and suppose that (i) to any two elements x, y of V there corresponds an element in V, called their sum,
- 36. 22 METRIC SPACES [2.2 which is denoted by x+y; and (ii) to any real number « and any x in V there corresponds an element in V, called the scalar product of « and x, which is denoted by ax. (Thus addition is a function on V x V into Vand scalar multiplication is a function on R'x V into V.) Suppose also that the following conditions are satisfied. () x+y = y+; (2) x+(y+z) = (x+y)+23 (3) there is an element 0 € V (the zero element) such that x+0 = x for allx eV; (4) to every element x€V there corresponds an element —xeV such that x+(—x) = 9; (5) a(Bx) = (@B)x; (©) 1x =x; (7) (@+A)x = ax+ Bx; (8) a(x+y) = axtay. Then V (together with the operations of addition and scalar multi- plication defined in V) is called a vector (or linear) space over R', or a real vector space. If the scalar product is formed with complex numbers « and the same properties (1)-(8) hold, then V is said to be a vector space over Z, or a complex vector space. The simplest example of a real vector space is R” with addition and scalar multiplication defined by (ay +25 Xn) +O, ---s Ra) = + Vv «+s Xn tPn), A(X, -++9 Xn) = (WX, «+, WXp)- A more sophisticated example is Bla, b]; here f+g and «fare defined by F+9® =f/@%)+s@), CO=7%0) @<x< bd). We shall assume the elementary algebraic properties of vector spaces. Some of these are given in exercise 2(b), 1. Definition. Let V be a vector space. The real valued function || .|| on V is called a norm on V, and V is called a normed vector space, if N1. ||x|| > 0 for all x € V and ||x|| = 0 if and only if x = 0; N2. |x+y| < ||| +ly] for all x, ye V; N3. ||ax|| = |e|||x| for every number « and all x € V. It is easy to see that, in a normed vector space, a metric p may be defined by A(x, y) = |]x—y]-
- 37. 2.2] NORMS 23 In fact, the metrics in illustrations (ii)-(v) of §2.1 all arise from norms. For instance in R” the usual norm is given by [|x] = V@i+... +27) and in B[a, b] |fl = sup IfC0)]. arc However, a metric on a vector space need not be derivable from a norm (see exercise 2(6), 2). More specialized than normed vector spaces are inner product Spaces. Definition. Let V be a complex normed vector space and suppose that, with every ordered pair (x,y) of elements in V there is associated a complex number, denoted by x.y, and that the following conditions are satisfied. Il. x.y = y.x for all x,y eV; 12. (ax+fy).z = a(x.z)+f(y.z) for all numbers a, 8 and all x,y, ZEV; 13. x.x = |x|? for all xe V. Then x.y is called the inner product of x and y and V is called a complex inner product space. If V is areal normed vector space, x.y is real and 11-13 hold, then V is called a real inner product space. Note that, in this case, I1 reduces toe ya yer: It is shown in exercise 2(b), 4 that no more than one inner product can be associated with a given norm. Once again R” provides the most obvious illustration. The inner product for the usual norm is defined by XV = Mite + kane The inequality (2.13) can now be written as x.y < |x] [vl and is a particular case of the following result. Theorem 2.21. (Cauchy’s inequality.) If V is a real or complex inner product space, then, for all x, y € V, Equality holds if and only if one of x, y isa scalar multiple of the other.
- 38. 24 METRIC SPACES [2.2 Proof. Since 0. = 0 (see exercise 2(b), 6), we may assume that x + 0. Take yee Then 0 < ||Ax+yl/? = (Ax+y).(Ax+y) = ew oben elie — le¥lP ya 9 1: — Fak bp 2te Moreover ||£x+yl| = 0, ie. Ax+y = 6, if and only if y = ax for some number «. | Exercises 2(5) 1. Show that the following identities hold in a vector space: (i) Ox = 0; (ii) (-I)x = —x; (iii) 20 = 0; (iv) if x—y denotes x+(—y), then Ey) a ee YX 2. Prove that the discrete metric 7 on R? (see illustration (i) of § 2.1) cannot be derived from a norm. 3. Two norms on a vector space V are equivalent if they give rise to equivalent metrics, Prove that the norms |.|, and |], are equivalent if and only if there exist strictly positive constants A, 4 such that Axl, < Ixle < allt, for all x € V. (Contrast exercise 2(a), 7.) 4. Show that if, in a normed vector space, an inner product exists, then it is unique. 5. The space T consists of the sequences x = (x,, X2, ...) of complex numbers such that &|x,|? converges. Verify that p given by oss) = ( beam aalt) Ge yeT) is a metric on 7; that the metric arises from a norm; and that to this norm there corresponds the inner product © X.Y = YD XnTne n=l 6. Show that in an inner product space x.0=0.x=0. 7. Prove that, if V is an inner product space, then, for all x, ye V, z [x+y]?+ xy]? = 2(1x|?+ | 719- Hence give an example of a normed vector space which is not an inner product space.
- 39. 2.2] NORMS 25 8. The sequences (@n), (Yn) in the normed vector space V converge to x,y Tespectively, and the numerical sequences (@n), (2) converge to a, 2 respectively. Show that %mXn+ Bn¥n > &x+ By. Prove also that, if V possesses an inner product, then Xn-Yn > X.Y. 9. Alternative definition of inner product space. Let V be a complex (real) vector space and suppose that, with every ordered pair (x, y) of elements in V, there is associated a complex (real) number x.y such that I1, I2 are satisfied and 13’. for all xe V, x.x > O and x.x = 0 if and only ifx = 0. Show that ||x|| = ./(x.x) is a norm on V. 2.3. Open and closed sets In this section we consider subsets of a fixed metric space (X, p). We shall use the symbol X for the metric space as well as the set without reiterating that p is specified. Let ae X and let r be a positive real number. The set of points x such that p(a, x) < ris called the open ball with centre a and radius r; we shall denote it by B(a;r). In R', B(a;r) is the open interval (a—r, a+r). In R? and in Z an open ball is usually called an open disc. — Definition. Given a subset E of X (which may be X itself), the point cé€X is said to be a limit point of E if, for every € > 0, there is an x €Esuch that 0<p(ce,x) <6, i.e. B(c; €)—{c} contains a point of E. A limit point c of E can also be defined by either of the following two conditions: fj (i) Every open ball B(c; ¢) contains infinitely many points of E. (ii) There is a sequence of points x, ¢ such that x, +c and X, > casn>o. It is clear that a point c with properties (i) or (ii) is a limit point ofE. Conversely, if c is a limit point of E, then, for every n, there is a point X, € E such that 0 < p(c, x) < In and so (ii) is satisfied. Also (ii) implies (i). The conditions (i) or (ii) show that a finite set cannot have a limit oint. An infinite set may or may not have a limit point. For example the subset {1, 2, 3, ...} of R! has no limit point. It is important to note that a limit point of a set Z may or may not belong to £. For instance if, in R', E is the interval (0, 1), then every point in the interval [0, 1] is a limit point of E.
- 40. is Nu open" tera SSS SRC IGAsal) eGo Ras 26 METRIC SPACES [2.3 A point c of E is sai isolated poi) E if it is not point of E. Thus, if c is an isolated point of £, there isa é > Osuch laaroagat Somes . . that B(c; 6) contains no point of £ other than c. Definition. Given a set E < X, the point c € Eis said to be an interior point of E if there is a 8 > 0 such that B(c; 8) © E. Illustrations Ai All points of an open ball B(a; r) are interior points of B(a; r). For suppose that c € B(a; r), so that p(a, c) = s < r. If 6 = r—s and p(c, x) < 4, then pa, x) < pla,c)+plc, x) < st+d=r and so x€B(a;r). Therefore B(c; 6) < B(a;r), i.e. c is an interior point of Ba; r). Gi) X = Ri. Let E = [0, 1). The set of limit points of E is [0,1]; the set of interior points of E is (0, 1). —~TLet FE be the set of rational points of Rt. Every point of R* is a limit point of E. E has no interior points. Gi) X = Z. Let E be the annulus consisting of the points z such that 1 < |z| < 2. The set of limit points is the set of z such that 1 < |z| < 2. The set of interior points is E. Let E be the set of points x+iy such that y = —1 and 0 < x < 1. The set of limit points is E. There are no interior points. When the space X has no isolated points (as is the case with R” an then an interior point of a subset E of X is also a of E. But if X is, for instance, the space consisting of allth (andp is the usual metric in R¥), then every subset E of X is without wits limit points although all the points of E are interior points of E. HS. ie —— Definition. (i) A set F in X is said to be closed if F contains all its limit points. (ii) A set G in X is said to be open if every point of G is an interior point of G. a Y Illustrations @ Every open ball in a metric space is open. A set {x|p(a, x) < r} is closed (see exercise 2(c), 2) and is consequently called the closed ball with centre a and radius r. = - 7 Any set without limit points is closed; in particular, every finite set is closed. V/ (ii) In R", every open interval is open and every closed interval is closed. o/ (ii) In R', the interval [0, 1) is neither open nor closed; the set of rational points is neither open nor closed, (iv) In Z, the annulus defined by 1 < |z| < 2 is open; the real axis (im z = 0) is closed; the set of points z = (1+i)/n(m = 1, 2, ...) is neither open nor closed.
- 41. 2.3] OPEN AND CLOSED SETS ZT (v) Let (Ca, 5], p) be the metric space of real functions continuous (and so bounded) on the interval [a, 5], in which PG 8) = oon |f@)—-g@)|- a<r< Then the set E, of f such that inf f(x) >0 a<x<b is open and the set E, of f such that f@=1 is closed. (vi) In any metric space the empty set and the whole space are both open closed. In R” and Z these are the only sets that are both open and closed (theorem -$:37, corollary). In an arbitrary metric space there may be others. For instance in the space of integers (with the usual metric of R) every set is both open and closed. The question is taken up again in §3.3. (Wii) The property of being open or closed (or neither) is not intrinsic to a given set £; it is relative to t i i subset metric p wil i i . For instance [0, 1) is neither open nor closed as a subset of Rt equipped with the usual metric, but [0, 1) is closed as a subset of (—1, 1) with the same metric; and as a subset of (R!, 7), where is the discrete metric, [0, 1) is both open and closed. A convenient characterization of closed sets is embodied in the following result. Theorem 2.31. A necessary and sufficient condition for the set F to be closed is that, whenever (x,,) is a convergent sequence of points in F, lim x, € F. no Proof. First suppose that F is closed and that x, — x, where X, € F for every n. If x,, = x for all sufficiently large n, then, a fortiori, x € F; otherwise x is a limit point of F and again xe F. Next, let F have the property described in the theorem. If F has no limit points, then F is closed. If F has a limit point x, then there is a sequence (x,) of points in F such that x, > x, and so xe F. | Theorem 2.32. (i) The set G in X is open if and only if G’ (= X—G) is closed. (ii) The set F in X is closed if and only if F’ (= X—F) is open. Proof. We need only show that (a) if G is open, then G’ is closed and (6) if Fis closed, then F’ is open. (a) Let G be open. If G’ has no limit points, then G’ is closed. If G’ has a limit point c, then every open ball B(c; ¢) contains points of G’. Hence c cannot be an interior point of G and, since G is open, c ¢ G. Thus ce G’. We have therefore shown that G’ contains all its limit points.
- 42. 28 METRIC SPACES [2.3 (b) Let F be closed. If F’ is empty, then F’ is open. If F’ is not empty, let x be any point of F’. Since Fis closed and x ¢ F, it follows that x is not a limit point of F. Hence there is a B(x; 5) free of points of F, i.e. there is a B(x; 6) < F’. Thus the arbitrary point x of F’ is an interior point of F’. | We have stressed the fact that a given set is open or closed (or neither) only with respect to the metric space in which it is embedded. Ttis therefore of interest to note that, if p and o are equivalent metrics on X, then the metric spaces (X, p) and (X, 7) have the same open and the same closed sets. The converse also holds. Both statements follow easily from the last two theorems. The next theorem concerns collections of open or closed sets. These collections may be finite or infinite and need not be countable. Theorem 2.33. (i) The union of any collection of open sets is open. ( (ii) The intersection of any collection of closed sets is closed. Proof. (i) Let Y be a collection of open sets G. If xe U G, then Geg there is a member of Y, say G*, such that x € G*. Since G* is open, x is an interior point of G*, i.e. there is a 6 such that B(x; 6) < G* and so also Bex; 6) < UG. Geg Hence x is an interior point of U G. This shows that U G is open. Geg Geg (ii) Let F be a collection of closed sets F. We could show directly that Al. F contains all its limit points; but, to illustrate the use of complements in proving ‘dual’ results for open and closed sets, we use (i) as follows. Since F’ is open, whenever Fe F, by (i), COED) ts umn FeF FeF is open. Hence, by theorem 2.32, () Fis closed. | FeF Theorem 2.34. (i) The intersection of a finite collection of open sets is open. (ii) The union of a finite collection of closed sets is closed. The proof is left to the reader.
- 43. Reeth NSS TR AB i Poses => CO. 10 dered. 2.3) OPEN AND CLOSED SETS 29 Exercise. Give examples of (i) a sequence of open sets whose intersection is not ithe AA qe! open and (ii) a sequence of closed sets whose union is not closed. wu) ia ’ -4 n Let (Y, 7) be a subspace of the metric space (X, p). Then (Y, ) has open and closed sets and, although a set which is open (closed) in Y is, of course, generally not open (closed) in X, the relationship be- tween open and closed sets in X and in Y is very simple. Theorem 2.35, Let (Y, 0) be a metric subspace of the metric space (X, p). Then a set is open in (Y,«) if and only if it is of the form Y 0 G, where G is open in (X, p). A similar result holds for closed sets. Proof. First, if G is open in X, then clearly every point of Y n G is an interior point of Y 0 Gin Y_ cl Next, let E be open in Y. Then, for every x € E we can find a 6, such that y¢ E whenever ye Y and o(x, y) < 6,. If now B(x; 6,) => (0,!) denotes an open ball in X, the set G —1UPBG6,) zeE is openin X¥ and E = Yn G. The result for closed sets may be deduced by the method of taking complements. | Corollary. If Y is open in X, then a subset of Y is open in Y if and only if it is open in X. If Y is closed in X, then a subset of Y is closed in Y if and only if it is closed in X. Definition. Let E be any subset of X. (i) The interior of E, denoted by E°, is the set of interior points of E. (ii) The closure of E, denoted by E, is the union of E and the set of limit points of E. Thus E£ is the set of x such that every B(x; €) contains at least one point of E or, equivalently, x € £ if and only if x is the limit of a sequence of points in E. For alternative definitions of the notions of interior and closure see exercise 2(c), 6. Theorem 2.36. For any set E in X, E° is open and E is closed. Proof. If E° is empty, it is open. If E° is not empty, let x be any point of £°. Then there is a B(x; 6) ¢ E. Every ye B(x; 6) is an interior point of B(x; 6) and so of E. Hence B(x; 6) < E° and there- fore x is an interior point of E°.
- 44. 30 METRIC SPACES (2.3 It may be shown directly that £ is closed. The result also follows from the identity (BY = (EY which is easily proved. (See exercise 2(c), 7.) | Open and closed sets could also have been defined in terms of interiors and closures: G is open if and only if G° = G; Fis closed if and only if F = F. Definition. A frontier point of a set E in X is a point c such that every open ball B(c; €) contains at least one point of E and at least one point of E'. The set of all frontier points of E is called the frontier of E and is denoted by fr E. Clearly E and E’ have the same frontier. Since a frontier point of E is a point of E which is a limit point of E’ ora point of E’ which is a limit point of Z, it follows that fr E = E-E°. (2.31) Illustrations (i) Let E be the interval [0, 1) in R*. Then E° = (0, 1), E = [0, 1] and frE = {0, 1}. (ii) Let (X,) be the space of integers with the metric induced by R. If E= B¢c; 1) = {c}, then E° = E = E. Note that E is not the closed ball {x|p(x, c) < 1} = {e-1, c, c+ 1}. Exercises 2(c) 1. Show that every infinite set Y may be equipped with a metric p which is such that X has a limit point in CX, p). 2. Prove that, in any metric space (X, p), the closed ball {x|p(a, x) < r}is closed. 3. Prove that a bounded, closed set of real numbers contains its supremum and infimum. . 4. Let E be an arbitrary set in a metric space. Show that the set E* of limit points of E is closed. 5. Show that in R" with the usual metric any collection of disjoint open sets is countable. Is this true for an arbitrary metric space? 6. (i) Show that E° is the union of all open sets contained in E and so is the ‘largest’ open set contained in E. (ii) Show that E is the intersection of all closed sets containing E and so is the ‘smallest’ closed set containing E. 7. Prove that, for any subset E of X, E°ufr Ev (E)° = X and @y =): (The set (E’)° is called the exterior of E.)
- 45. 2.3] NOTES 31 8. Prove that frE= EnE’ and that fr E is closed. 9. (i) Prove that Bice Ub. =hy UU Be but that, for an infinite collection & of sets E, generally the relation UE >UE Ecé Ecé only holds. (ii) Prove that, for any collection & of sets E, NES EF Ecé Ecé Construct an example to show that identity need not hold even for a finite collection. Gii) Deduce the results for interiors corresponding to (i) and (ii). 10. The metric spaces (X, p), (X, 7) have equivalent metrics. Show that every subset E of X has the same limit points and interior points in (X, p) and CX, c). 11. If EZ is a set in a metric space and the set D is such that DSH S-D: then D is said to be dense in E. (For instance the set of rational numbers is dense in R'.) Show that, if C is dense in D and D is dense in £, then C is dense in E. NOTES ON CHAPTER 2 §2.1. The idea of a metric space is due to M. Fréchet. In his doctoral thesis, published in 1906, he examined various sets of axioms for a metric. The set M1-—M:S3 is the one which experience showed to be the most fruitful. The inequality (2.12), which is equivalent to the triangle inequality for the usual metric on R”, is a particular case of Minkowski’s inequality (3, oro)” < (Sa) (Se) in which a,, b; > 0 (i = 1, ... n) andr > 1. (In (2.12) the a;, b; may be negative, but the step from non-negative a;, b; to arbitrary ones is trivial.) The inequality (2.13), which we call Cauchy’s inequality in theorem 2.21, is also known as Schwarz’s inequality (or even the Cauchy-Schwarz inequality). It is a particular case of Hélder’s inequality. n n p/n 1/q E ab, < (3 at) (3) : i=l i=1 i=1 where a;,b; > 0 (Gi =1,..-,”), p> 1 and p++q = 1. Both Hdlder’s and Minkowski’s inequalities may be generalized in a variety of ways. (See Hardy, Littlewood and Pélya, Inequalities.)
- 46. 3 CONTINUOUS FUNCTIONS ON METRIC SPACES 3.1. Limits The notion of continuity is based on that of a limit and we first define the limit of a function whose domain and range lie in arbitrary metric spaces. Definition. Let (X, p) and (Y, o) be metric spaces. Let f be a function with domain E < X and with range in Y. Suppose also that x is a limit point of E and y, € Y. We then say f&) > a x>X or lim f(x) =% if, given € > 0, there is a 3 > O such that = o(f(x),¥0) < € whenever xe E eli ie Xo) < 6. Notes. (i) It is irrelevant whether x, belongs to'the domain E of f or not. mg : (ii) When X¥ = Y = R° (or Z) the above definition reduces to the standard definition of the limit of a real (or complex) function. Exercise. Show that a function cannot converge at a point to more than one limit. Mlustrations @ X= Y=R @) With E = R} lima? = 2% (K = 1, 2,...). IX (6) With E = R‘— {0}, lim x sin (1/x) = 0, lim sin (1/x) does not exist. x0 z0 (c) Let E be the set of rational numbers and let f be defined on E by the ee f(pla) = Va when p, g are coprime integers and g > 1. We shall show that, for every c € R}, lim f(x) = 0. ae Given ¢ > 0, let q be an integer such that 1/g, < ¢. Denote by Q, the set of rational numbers piq with g < qo. We note that, in any finite interval, there is only a finite number of members of Q,. Hence there is a 6 > 0 such that the intervals (c—6,c) and (c,c+9) contain no members of Qy. It follows that, whenever 0 < |c—p/q| < 6, o<s(2)=teh ce q} 4 4%
- 47. 3.1] LIMITS 33 Gi) X = R*, FE = R*-{(0, 0)}, Y= R. Let g, be given by Ga x HOY) = ae x? Then lex | = xl eas lal and lim __g,(x, y) = 0. , (x, v)>(0, 0) . Let gy be given by als) == ey Then te paar 4 B(x, ¥) does not exist. For, if y _= mx,we have (for all x + 0), Be 1 82%, mx) = a + meet +m? and so g, takes all values between 0 and 1 in any annulus 0 < x?+y? < 62, ew (iii) Let E - X = ¥ = C(O, 1] (the set of real functions continuous on [0, 1]). The metric p is, as is usual on C[0, 1], given by Pb. $2) = sup |6@)—¢,(0. The function A from C[0, 1] to C[0, 1] is given by A(¢) = ¥, where Ho = [$00 ar. If, for instance, ¢» is given by ¢o(t) = #2, then lim A(¢) = Wo, where yro(t) = 42°. od. For, if p($o, ¢) < ¢, then, for0 < ¢ < 1, rea fo ar| = | [eran ar| < [ir-senlar <e 0 ho-YO| = and therefore p(yro, y) < €. A sequence 4, ds, a; ... may be regarded as a function on the set {1, 4, 4, ...}. It follows that the notion of a sequential limit may be expressed by means of the notion of a functional limit. What is less obvious, and more interesting, is that the opposite statement is also true. Theorem 3.1. Let (X,p) and (Y,0) be metric spaces. Let f be a function with domain E < X and with range in Y and suppose that xo is a limit point of E. Then f(x) > yo as x > Xp if and only if, for every sequence (Xp) in E—{xo} such that x, > Xo, f(Xn) > Yo asin > 00. Proof. (i) Suppose that f(x) > Vo as x > Xo. Then, given ¢ > 0, there is a 6 > 0 such that o(f(x), Yo) < € whenever x € E and 0 < p(x, x9) < 9. Bsc
- 48. 34 CONTINUOUS FUNCTIONS ON METRIC SPACES [3.1 Now, if (x,,) is any sequence in E—{xo} such that x, > Xo, there isan integer m such that 0 < p(Xn, Xo) < & for n> Mm. Therefore o(f(Xn)s Yo) < € for n> Mm, ne S(%n) > Yo AS n> OO. (ii) Suppose that, whenever (x,) is in E—{x} and x, > Xo, F(%n) > Yo- If f(x) +> yo aS X > Xo, then there is an ¢ > O with the property that there is no 6 > 0 such that o( f(x), yo) < € whenever x € E and 0 < p(x, xo) < 6. Therefore, given any n, there is an x, € E such that 0 < (Xn, Xa) < In and o(f(%q), Yo) > & Thus there is a sequence (x,) in E—{xo} such that x, > x) and F(%n) +> Yo. This contradicts our original hypothesis and so f(x) > yo as X > Xp. | For functions with values in a normed vector space V there is, as might be expected, an algebra of limits. Let f; g be such functions defined on a subset E of X. Also suppose that E has a limit point x9 — L]) > Yoo B(x) > 2% aS X> Xp. When ¢, are real or complex valued functions on E and d(xa>a, Yx>h as x>X, then PODS) + VOQ)B(x) > 2Y0 + Azo. (3.11) If V possesses an inner product, then also F(X) .8(%) > Yo-Z05 (3.12) and if Vis R? or Z, F)/g(X) > yolZ0 (3.13) provided that z) + 0. These results may be proved directly or they may be deduced, by means of theorem 3,1, from the corresponding results on sequences (exercise 2(b), 8). Exercises 3(a) 1. The real valued function fon R?—{(0, 0)} is defined by x*y fy) = Taya Show that lim f(x, y) does not exist. (@, v)—>(0, 0) va
- 49. 3.1] LIMITS 35 Za Let (X, Pp), (Y, ©) be metric spaces. Let f be a function with domain E < X and with range in Y, and suppose that x, is a limit point of E. Show that, if lim f(x,) exists for every sequence (x,) in E—{x9} such that x, > Xp, then n>o lim f(x) exists. 22, 3. Let (X, p), (Y, 0) be metric spaces, and let E be a subset of X, x, a limit point of E, yo a point of Y. Show that, if the function f: E> Y is such that f(x) > yo as x > Xo, then yp € f(E); and that if, in addition, fis injective, then y, is a limit point of f(£). 4. Let (X, p), (¥, o), (W,7) be metric spaces and let D, E be subsets of X, Y respectively. Leth = go f: D > Wbe the composition of the functions f:D > Y, where f(D) © E, and g: E> W. Also (i) Xp is a limit point of D and f(x) > yo as x > X9; (ii) yo is a limit point of E and g(y) > wo as y > yo. Show that, whenf is injective (so that, by 3, yo is automatically a limit point of E), then h(x) > wo as x > x9. Construct an example in which A(x) +> wo. 5. Prove that the statements (3.11)-(3.13) hold under the conditions given at the end of the section. 3.2. Continuous functions The definition of continuity which we shall now give is a natural generalization of the concept familiar in elementary analysis. > Definition. Let (X, p) and (Y, c) be metric spaces. Then the function f: X > Yis said to be continuous at the point xy of X if, given e > 0, there is a 6 > O such that o(f(x),f(%)) < € whenever p(x, Xo) < 6 (or f(B@%o; 6) = BFC); €))- If f is continuous at all points of a set, it is said to be continuous on. that set. The definition implies that f is continuous at a limit point x) of xt f) +f) _as_ x >; and that fis continuous at all isolated points of X (for if x, is an isolated point and 6 is sufficiently small, the only point x such that p(X, Xo) < Fis Xp itself). ~Let f be a function whose domain is a subset X, of a metric space (X, p). When considering the limits of f we are interested in the limit points of X, and these need not belong to Xj. However, for questions Pe <i} of continuity, only the points of the domain X, of f are relevant. re Since X,, endowed with the metric induced by (YX, p), is itself a metric space, it is therefore simplest in this context to treat X, as the 2-2
- 50. 36 CONTINUOUS FUNCTIONS ON METRIC SPACES 3.2 fundamental space. This is, in fact, what we did in the definition of continuity. Mlustrations. The function f of illustration (i)(c) on p. 32 whose domain is de set of rational numbers, is discontinuous everywhere. If f,, with domain R* is ety f(x) when x is rational, (x) = cone A 0 when x is irrational, then f; is continuous at all irrational points and discontinuous at all rational points. : : . } The function h: C[0, 1] > C[0, 1] of illustration (iii) on p. 33 is continuous on C[0, 1]. This follows from the inequality PCACA,), hs) < PCr» $2) which holds for all ¢,, 6. of C[0, 1]. : Let (X, p) be any metric space, let a be a fixed point of X and let the function g: X > R' be defined by the equation a(x) = pCa, x). Since P(@, x1) —P(@, x2) < P(%2, X) and P(@, X2)— P(A, X31) < PC; Xa), le@—g@)| = |p, x)—P(@, x2)| < PO, x2) for all x,, x, in X. Hence g is continuous on X. Algebraic operations on continuous functions with values in a normed vector space again yield continuous functions. If the functions f,g: X > V and the functions ¢, y: X > R! (or Z) are continuous at xo, then ¢f+yeg is continuous at x; so is f.g if V is an inner product space, and //gis continuous at x» if Vis R! or Z and g(xp) + 0. All this is obvious when xy is an isolated point of ¥ and otherwise follows from the corresponding results on limits. In the next theorem we return to functions with arbitrary ranges. The result contrasts with exercise 3(a), 4. Theorem 3.21. Let (X,p), (Y,o), (W,7) be metric spaces and let h = gof: X > W be the composition of the functions f: X > Y and g: Y> W. If fis continuous at x» and g is continuous at yy) = f(x), then h is continuous at x». Proof. Given e, there is an 7 such that 7(g(y); 8(Yo)) < € whenever o(y, Yo) < 7. There is now a 6 such that o(f(x), f(%)) < 7 whenever p(x, x) < 6
- 51. 3.2] CONTINUOUS FUNCTIONS Kf and so @ is also such that (A(x), h(X)) < € whenever p(x, X9) < 6. | It is important to note that the inverse function of a bijective continuous function is generally not continuous. A simple example ‘is thefunction with domain Y = [0, 1) u [2, 3)and range Y = [0, 2] defined by 0) [x for0 <x <1, x) = x= for 2x < 3) If Xand Y have the metrics induced by R!, then fis continuous on X. However f1, which is given by y for0 < y <1, TO) = f (p41 forl < y < 2, is discontinuous at the point 1. For other examples see exercises 3(b), 4 and 5. The concept of continuity can be formulated in a variety of ways. We shall next consider equivalent definitions, first for continuity at a point and then for continuity on the domain of the function. Theorem 3.22. Let (X,p) and (Y,) be metric spaces. A necessary and sufficient condition for the function f: X > Y to be continuous at the point Xp is that x, > Xo implies f(Xn) >f(Xo)- The proof is similar to that of theorem 3.1. The theorem may also be deduced from theorem 3.1, but slight complications arise, since members of the sequence (x,) may now be Xo. Theorem 3.22 shows that the introduction of equivalent metrics in the domain or the range of a function does not affect the property “Of continuity: if p, , are equivalent metrics on X and o, oj are- equivalent metrics on Y, and if f: X > ¥ is continuous at x» with Tespect to p and o, then fis also continuous at xo with respect fo p; and 0}. —_ Theorem 3.23. Let (X,p) and (Y,) be metric spaces. Each of the following conditions is necessary and sufficient for the function f: X > Y to be continuous on X: (i) Whenever G is open in Y, then f-(G) is open in X. (ii) Whenever F is closed in Y, then f(F) is closed in X. Proof. (i) First suppose that f is continuous on X and that the set G is open in Y.
- 52. ¢ONTINUOUS FUNCTIONS ON METRIC SPACES B.2 If the set f-1(G) is empty then it is also open. If f(G) is not empty, let x) be any one of its points. Then f(x) € G and, since G is open, there is an ¢ such that yeG whenever o(y,f(%)) < & But, since f is continuous at Xo, there is a 6 such that o(f(x),f(%)) < € whenever p(x, Xo) < 6. Hence f(x)€G whenever p(x, x9) < 4, ie. xef-(G) whenever p(x, Xo) < 6. This shows that x) is an interior point of f(G). It follows that f-(G) is open in X. Next suppose that, whenever G is open in Y, then f(G) is open inX. Let x) be any point of X and take any positive e. The open ball B(f(%); €) is open in Y and so, by hypothesis, the set f{B( f(x); ©)} is open in X. This set also contains x, and therefore there isa d > 0 such that ef-{B(f(%); )} whenever p(x, Xo) < 6, ne F(x) € BU); ©) whenever p(x, Xo) < 4, Le, o(f(x), f(%o)) < € whenever p(x, X) < 6. Thus fis continuous at x) and so on X. (ii) This part follows from (i) by the method of complements since, if E is any subset of Y, XE) nfE)= @ and f(E)u fe) =X. | In the last theorem the inverse images may not be replaced by direct images. It is not true that /(S) is open (closed) in Y whenever Sis open (closed) in ¥. The function defined on p. 37 will illustrate this (exercise 3(b), 6) but it is instructive to employ also more common- place examples. For instance the function f with domain R! and range R! given by AO = Hee av maps the open interval (0, 1) onto (0, 3]. Next, let g, with domain (— ©, 00) and range [—4, 4] be defined by x ex g(x) = This function maps the closed interval [1, 00) on the interval (0, 4] which is not closed in [—4, 4].
- 53. 3.2] CONTINUOUS FUNCTIONS 39 We have previously mentioned that a bijective continuous function F: (X, p)> (Y, ¢) need not be such that f: (Y, c) > (X, p) is con- “tinuous. When /— is continuous, fis called a homeomorphism. (Note that then f~ is also a homeomorphism.) Two metric spaces are said to be homeomorphic if there exists a homeomorphism on one into the other. If f:(X,p)—>(Y,¢) is a homeomorphism, then, by theorem 3.23, f maps open (closed) sets in X into open (closed) sets in Y and there is a bijection between the open (respectively closed) sets in the two spaces. A bijective function f: (X, p) > (Y, 2) such that oF)» f&%2)) = PP 2) for all x;, x, € X is called an isometry. It is plain that an isometry is a homeomorphism. Two metric spaces are called isometric if there exists an isometry on one into the other. Such spaces have identical metric properties although their elements may be of entirely different kinds. For instance R} is isometric with the subspace A of C[0, 1] consisting of the functions ¢, (— «0 < A < ) defined by d(x) = Ax O<x< 1). Another example of an isometric pair is provided by R®, Z. Linear functions. We briefly consider a class of functions which plays an important part in modern analysis. Definition. Let V, W be two real (complex) normed vector spaces. A function f: V > W is called linear if S (GX, + HX) = Of (Xr) + of (2) for all x, X, € V and all real (complex) numbers 04, %». Clearly f(@) = 9, where the letter 0 on the left stands for the zero element in V and that on the right for the zero element in W. (It is not customary to use distinctive notations for the zero element and the norm in the two spaces.) Mlustrations (1) The function f: C[a, b] > R} defined by the relation £) = fo dt (pe Cla, b) is linear. : (2) The reader is probably acquainted with linear functions on one Euclidean space into another. We shall use three properties of these functions which are
- 54. 40 CONTINUOUS FUNCTIONS ON METRIC SPACES [3.2 proved in most books on linear algebra. (See, for instance, L. Mirsky, Intro- duction to Linear Algebra, chapters IV and V.) (i) The function f: R" + R” is linear if and only if it is of the form LG, 0103 Xm) = Vas v9 Yn)r where V=au%it. nent 21) Vn = An Xt... + AnmXm Va Qy1---Bym [%1 or Noa Sipe v a ONE Cell Vee The nx m matrix (a,;) is said to represent f. (ii) If the linear functions f: R” > R” and g: R" > R® have matrices A, B, then the linear function go f: R” > R? is represented by the matrix BA. ii) Let f: R"™ > R* be a linear function with matrix A. Then f is injective if and only ifm > mand A has (maximum) rank m. Also fis surjective (i.e. the range of fis the whole of R”) if and only if < mand A has (maximum) rank n. Thus, a necessary and sufficient condition for f to be bijective (so that f— exists and is again a linear function) is that m = n and A has maximum rank, i.e. det A + 0. When f is bijective, f-! has matrix A“. It is easy to see that a linear function on R” into R” is continuous at all points of its domain (see exercise 3(b), 8). A consequence of the theorem below is that any linear function either is everywhere continuous or is nowhere continuous. Ne Theorem 3.24. Let V, W be normed vector spaces. If the function y JS: V + Wis linear, then the following three statements are equivalent. Shi (i) fis continuous on V. ire) | (ii) There is a point x») € V at which f is continuous. (iii) ||f(%)||/|| || is bounded for x € V—{6}. pey 5 ja A a Proof. (ii) = (i). Take any ¢ > 0. Since f is continuous at xp, there is a 6 > 0 such that ||x— xl < 6 implies || f(x) —f(x)|| < «. Let c be any point of V and take any x such that ||x—cl| < 6. Then |(@-e+x)—xl| < 6 and therefore ||f(x-—c+x)—f(x)| < 6 ie. |f0C)—f(0)|| < €. Thus fis continuous at c. (i)+ (iii). Suppose that ||f(x)||/||x| is not bounded. Then, given any integer 7, there is a point x, € V—{6} such that ||f(x,,)]|/||xnl| > 7. If v, = (||x,||)-1x,, then ||v,|| = 1/n and _ If@nl, |Fe,)ll “nx > Hence fis not continuous at 0.
- 55. Another Random Document on Scribd Without Any Related Topics
- 56. “Your dress, too, Hilda, is lovely.” “Do you notice dresses, care about them?” “I think I do, sometimes; not in detail as a woman would, but in the blended effect of dress and wearer.” “I love beautiful dresses. I think this dress is beautiful. Have you noticed the line it makes from breast to hem, that long, unbroken line? I think that line the secret of elegance. In some gowns one sees one has visions of crushed ribs, don’t you think?” Odd listened respectfully, his mouth twisted a little by that same smile that he still felt to be painful. “And is not this lace gathered around the shoulders pretty too?” Hilda turned to him for inspection. “You will talk about your clothes, but you will not talk about yourself, Hilda.” Odd had put on his eyeglasses and was obediently studying her gown. “The lace is mamma’s. Poor mamma; I know she is lonely. It does seem hard to be left alone when other people are enjoying themselves. She has Meredith’s last novel, however. I began it with her. Mr. Odd, I am doing all the talking. You talk now.” “About Meredith, your dress, or you?” “About yourself, if you please.” “It has seemed to me, Hilda, that you were even less interested in me than you were in yourself.” Hilda looked round at him quickly, and he felt that his eyes held hers with a force which almost compelled her— “No; I am very much interested in you.” Odd was silent, studying her face with much the same expression that he had studied her gown—the expression of painfully controlled emotion. “There is nothing comparably interesting in me,” he said; “I have had my story, or at least I have missed my chance to have a story.” “What do you mean?” “Well, I mean that I might have made a mark in the world and didn’t.”
- 57. “And your books?” “They are as negative as I am.” “Yet they have helped me to live.” Hilda looked hard at him while she spoke, and a sudden color swept into her face; no confusion, but the emotion of impulsive resolution. Odd, however, turned white. “Helped you to live, Hilda!” he almost stammered; “my gropings!” “You may call them gropings, but they led me. Perhaps you were like Virgil to Statius, in Dante. You know? You bore your light behind and lit my path!” She smiled, adding: “I suppose you think you have failed because you have reached no dogmatic absolute conclusion. But you yourself praise noble failure and scorn cheap success.” “I didn’t even know you read my books.” “I know your books very well; much better than I know you.” “Don’t say that. I hope that any worth in me is in them.” “One would have to survey your life as a whole to be sure of that. Perhaps you do even better than you write.” “Ah, no, no; I can praise the books by that comparison.” His voice stumbled a little incoherently, and Hilda, rising, said with a smile— “Shall we dance?” In the terribly disquieting whirl of his thoughts, which shared the dance’s circling propensities, Odd held fast to one fixed kernel of desire; he must hear from Hilda’s lips why she had refused Allan Hope. An uneasy consciousness of Katherine crossed his mind once and again with a dull ache of self-reproach, all the more insistent from his realization that its cause was not so much the infidelity to Katherine as that Hilda would think him a sorry villain. Katherine seemed to be dancing and enjoying herself. She knew that his energy this evening was on Hilda’s account; he had claimed the responsibility for Hilda. Katherine would not consider herself neglected, of that Peter felt sure, relying, with perhaps a display of the dulness she had discovered in him, upon her confidence and
- 58. common sense. Outwardly, at least, he would never betray that confidence; there was some rather dislocated consolation in that. Hilda was a little breathless when he came to claim her for the second cluster of waltzes. It was near the end of the evening. “I have been dancing steadily,” she announced, “and twice down to supper! Did you try any of the narrow little sandwiches? So good!” “And you still don’t grudge me my waltzes?” “I like yours best!” she said, smiling at him as she laid her hand on his shoulder. They took a few turns around the room and then Hilda owned that she was a little tired. They sat down again on the sofa. “Hilda!” said Odd suddenly, “will you think me very rude if I ask you why you refused Allan Hope?” Hilda turned a startled glance upon him. “No; perhaps not,” she answered, though the voice was rather frigid. “You don’t think I have a right to ask, do you?” “Well, the answer is so evident.” “Is it?” Hilda had looked away at the dancers; she turned her head now half unwillingly and glanced at him, smiling. “I would not have refused him if I had loved him, would I? You know that. It doesn’t seem quite fair, quite kind, to talk of, does it?” “Not to me even? I have been interested in it for a long time. Katherine told me, and Mary.” “I don’t know why they should have been so sure,” said Hilda, with some hardness of tone. “I never encouraged him. I avoided him.” She looked at Odd again. “But I am not angry with you; if any one has a right, you have.” “Thanks; thanks, dear. You understand, you know my interest, my anxiety. It seemed so—happy for both. And you care for no one else?” “No one else.” Hilda’s eyes rested on his with clear sincerity.
- 59. “Don’t you ever intend to marry, Hilda?” Odd was leaning forward, his elbows on his knees, and looking at the floor. There was certainly a tension in his voice, and he felt that Hilda was scanning him with some wonder. “Does a refusal to take one person imply that? I have made no vows.” “I don’t see—“ Odd paused; “I don’t see why you shouldn’t care for Hope.” “Are you going to plead his cause?” she asked lightly. “Would it not be for your happiness?” Odd sat upright now, putting on his eyeglasses and looking at her with a certain air of resolution. “I don’t love him.” Hilda returned the look sweetly and frankly. “What do you know of love, you child? Why not have given him a chance, put him on trial? Nothing wins a woman like wooing.” “How didactic we are becoming. I am afraid I should really get to loathe poor Lord Allan if I had given him leave to woo me.” “I suppose you think him too unindividual, too much of a pattern with other healthy and hearty young men. Don’t you know, foolish child, that a good man, a man who would love you as he would, make you the husband he would, is a rarity and very individual?” Odd found a perverse pleasure in his own paternally admonishing attitude. Hilda’s lightly amused but touched look implied a confidence so charming that he found the attitude sublimely courageous. “I suppose so,” she said, and she added, “I haven’t one word to say against Lord Allan, except—“ She paused meditatively. “Except what?” Odd asked rather breathlessly. “He doesn’t really need me.” “Doesn’t need you! Why, the man is desperately in love with you!” “He needs a wife, but he doesn’t need me.” “You are subtle, Hilda.”
- 60. “I don’t think I am that.” “You are waiting, then, for some one who can satisfy you as to his need of you?” “I shall only marry that person.” Hilda jumped up. “But I’m not waiting at all, you know. Dansons maintenant! Your task is nearly over!” It was very late when Odd gave Hilda up to her last partner, and joined Katherine in a small antechamber, where she was sitting among flowers, talking to an appreciative Frenchman. This gentleman, with the ceremonious bow of his race, made away when Miss Archinard’s fiancé appeared, and Odd dropped into the vacated seat with a horrible sinking of the heart. The dull self-reproach was now acute, he felt meanly guilty. Katherine looked at him funnily— very good-humoredly. “I didn’t know you had it in you to dance so well and so persistently, Peter. You have done honor to Hilda’s ball.” “I hope I wasn’t too selfishly monopolizing.” “Oh, you had a right to a certain monopoly since, owing to you only, she came,” and Katherine added, smiling still more good- humoredly, “I am not jealous, Peter.” He turned to look at her. The words, the playful tone in which they were uttered, struck him like a blow. His guilty consciousness of his own feeling gave them a supreme nobility. She was not jealous. What a cur he would be if ever he gave her apparent cause for jealousy. The cause was there; his task must be to keep it hidden. “But suppose I am?” he said; “you haven’t given me a single dance.” Katherine’s smile was placid; she did not say that he had not asked for one. Indeed they had rarely danced together. “I think of going to England in a day or two, Peter,” she observed. “The Devreuxs have asked me to spend a month with them.” Peter sat very still. “A sudden decision, Kathy?”
- 61. “No, not so sudden. Our tête-à-tête can’t be prolonged forever.” “Until our wedding day, you mean? Well, the wedding day must be fixed before you go.” “I yield. The first part of May.” “Three months! Let it be April at least, Kathy.” “No, I am for May.” “It’s an unlucky month.” “Oh, we can defy bad luck, can’t we?” Katherine smiled. “If you go away, I shall,” said Odd, after a moment’s silence. “Why, I thought you would stay here and look after mamma—and Hilda,” said Katherine slowly, and with a wondering thought for this revealment of poor Peter’s folly. Peter then intended to heroically sacrifice his infidelity. That he should think she did not see it! “I am not over this beastly cold yet. A trip through Provence would set me right. I should come back through Touraine just at the season of lilacs. I am afraid I should be useless here in Paris. I see so little of your mother—and Hilda. Arrange that Taylor shall go for her after her lessons.” “I am afraid that mamma can’t spare Taylor.” Peter moved impatiently. “Katherine, may I give you some money? She would take it from you. Persuade her to give up that work. You could do it delicately.” “As I have told you, you exaggerate my influence. She would suspect the donor. She would not take the money.” “I could speak to your father; lend him a sum.” Katherine flushed. “It would make him very angry with her if he knew. And the lessons are a fixed sum; only a steady income would be the equivalent.” “Oh dear!” sighed Peter. He suddenly realized that of late he had talked of little else but Hilda in his conversations with Katherine. “When do you go to London, dear?” he asked.
- 62. “The day after to-morrow.” Katherine, above the waving of her fan, smiled slightly at his change of tone. “Will you miss me, Peter?” “All the more for being cross with you. It is very wrong of you to play truant like this.” “It will be good for both of us.” Katherine’s voice was playful, and showed no trace of the bitterness she was feeling. “I might get tired of you, Peter, if I allowed myself no interludes. Absence is the best fuel to appreciation. I shall come back realizing more fully than ever your perfection.” “What a sage little person it is! Sarcastic as well! May I write to you very often?” “As often as you feel like it; but don’t force feeling.” “May I describe châteaux and churches? And will you read my descriptions if I do?” “With pleasure—and profit. Let me know, too, how the book gets on. Can I do anything for you at the British Museum?” It struck Katherine that the change in their relation which she now contemplated as very probably definite might well allow of a return to the first phase of their companionship. A letter from Allan Hope which she had received that morning, though satisfactory in many respects, was not quite so from an intellectual standpoint. An intellectual friendship with Peter Odd was a pleasant possession for any woman, and Katherine perhaps, with an excusable malice, rather anticipated the time when Peter might have regrets, and find in that friendship the solace of certain disappointments from which Katherine had almost decided not to withhold him. “I shall try to keep you profitably yoked, then, even in London, shall I?” said Odd, in reply to an offer more generous than he could have divined. “Discipline is good for a rebellious spirit like yours. Don’t be frightened, Kathy. Go and look at the Elgin Marbles if you like. I shall set you no heavier task.” “They are so profoundly melancholy in their cellared respectable abode, poor dears! I know they would have preferred dropping to pieces under a Greek sky. A cruel kindness to preserve them in an
- 63. insulting immortality. The frieze especially, stretched round the ugly wall like a butterfly under a glass case!” Odd laughed with more light-heartedness than he had felt for some time. It rejoiced him to feel that he still found Katherine charming. There must certainly be safety in that affectionate admiration. “I won’t even ask you to harrow your susceptibility by a look at the insulted frieze, then; you must know it well, to enter with such sympathy into its feelings. Only you must write, Katherine. I shall be lonely down there. A daily letter would be none too many.” “I can’t quite see why you are exiling yourself. Of course, the weather here is nasty just now. I have noticed your cough all the evening. Come and say good-bye to-morrow. I shall be very busy, so fix your hour.” “Our usual hour? In the morning?” “You will not see Hilda then.” “Hilda has had enough of me to-night, I am sure. You will kiss her au revoir for me.” Odd felt a certain triumph. Katherine’s departure could be taken as a merciful opportunity for makeshift flight. After a month or two of solitary wrestling and wandering, he might find that the dubiously directed forces of Providence were willing to help one who helped himself. His mind fastened persistently on the details of the suddenly entertained idea of escape from the madness he felt closing round him. The disclosure of his passion for Hilda stared him in the face. And how face the truth? A man may fight a dishonoring weakness, but how fight the realization that a love founded on highest things, stirring highest emotions in him, had, for the first time, come into his life, and too late? A love as far removed from the wrecking passion of his youth as it was from the affectionate rationality of his feeling toward Katherine; and yet, because of that tie, drifted into from a lazy indifference and kindness for which he cursed himself, capable of bringing him to a more fearful shipwreck.
- 64. Hilda’s selflessness was rather awful to the man who loved her, and gave her a power of clear perception that made sinking in her eyes more to be dreaded than any hurt to himself. And Peter departed for the South without seeing her again.
- 65. A CHAPTER IX N April sky smiled over Paris on the day of Odd’s return. A rather prolonged tour had tanned his face, and completely cured his lungs. He expected to find Katherine already in Paris; her last letters had announced her departure from a Surrey country house, and had implied some anxiety in regard to a prolonged illness of Mrs. Archinard’s. Katherine had written him very soon after their parting, that the Captain had gone on a yachting trip in the Mediterranean, and that she knew that he had left Hilda with money, so Peter need not worry. Peter had seen to this matter before leaving Paris, and had approved of the Captain’s projected jaunt. He surmised that her father’s absence would lighten Hilda’s load, and hoped that the sum he placed in the Captain’s hands—on the understanding that most of it was to be given to Hilda—but from her father, would relieve her from the necessity for teaching. Peter called at the Rue Pierre Charron early in the afternoon, but the servant (neither Taylor nor Wilson, but a more hybrid-looking individual with unmistakable culinary traces upon her countenance) told him that Mademoiselle Archinard had not yet arrived. Madame still in bed “toujours souffrante,” and “Mademoiselle ‘Ilda”—Odd had hesitated uncomfortably before asking for her—was out. “Pas bien non plus, celle-là,” she volunteered, with a kindly French familiarity that still more strongly emphasized the contrast with Taylor and Wilson; “Elle s’éreinte, voyez-vous monsieur, la pauvre demoiselle.” With a sick sense of calamity and helplessness upon him, Odd asked at what hours she might be found. All the morning, it seemed “Il faut bien qu’elle soigne madame, et puis elle m’aide. Je suis seule et la besogne serait par trop lourde,” and Rosalie also volunteered the remark that “Madame est très, mais très exigeante, nuit et jour; pas moyen de dormir avec une damê comme celle-là.”
- 66. Odd looked at his watch; it was almost five. If Hilda had kept to her days he should probably find her in the Rue d’Assas, and, with the angriest feelings for himself and for the whole Archinard family, Hilda excepted, he was driven there through a sudden shower that scudded in fretful clouds across the blue above. He was none too soon, for he caught sight of Hilda half-way up the street as they turned the corner. The sight of him, as he jumped out of the cab and waylaid her, half dazed her evidently. “You? I can hardly believe it!” she gasped, smiling, but in a voice that plainly showed over-wrought mental and physical conditions. She was wofully white and thin; the hollowed line of her cheek gave to her lips a prominence pathetically, heartrendingly childlike; her clothes had reached a pitch of shabbiness that could hardly claim gentility; the slits in her umbrella and the battered shapelessness of her miserable little hat symbolized a biting poverty. “Hilda! Hilda!” was all Odd found to say as he put her into the cab. He was aghast. “I am glad to see you,” she said, and her voice had a forced gayety over its real weakness; “I haven’t seen any of my people for so long, except mamma. An illness seems to put years between things, doesn’t it? Poor mamma has been so really ill. It has troubled me horribly, for I could not tell whether it were grave enough to bring back papa and Katherine; but Katherine is coming. I expected her a day or two ago, and mamma is much, much better. As for papa, the last time I heard from him he was in Greece and going on to Constantinople. I am glad now that he hasn’t been needlessly frightened, for he will get all my last letters together, and will hear that she is almost well again. And you are here! And Kathy coming! I feel that all my clouds are breaking.” Odd could trust his voice now; her courage, strung as he felt it to be over depths of dreadful suffering, nerved him to a greater self- control. “If I had known I would have come sooner,” he said; “you would have let me help you, wouldn’t you?”
- 67. “I am afraid you couldn’t have helped me. That is the worst of illness, one can only wait; but you would have cheered me up.” “My poor child!” Odd inwardly cursed himself. “If I had known! What have you been doing to yourself, Hilda? You look—“ “Fagged, don’t I? It is the anxiety; I have given up half my work since you left; my pictures are accepted at the Champs de Mars. We’ll all go to the vernissage together. And, as they were done, I let Miss Latimer have the studio for the whole day. That left me my mornings free for mamma.” “Taylor helped you, I suppose?” “Taylor is with Katherine. She went before mamma was at all ill, and indeed mamma insisted that Katherine must have her maid. I was glad that she should go, for she has worked hard without a rest for so long, and, of course, travelling about as she has been doing, Katherine needed her.” There was an explanatory note in Hilda’s voice; indeed Odd’s silence, big with comment, gave it a touch of defiance. “It made double duty for Rosalie, but she is a good, willing creature, and has not minded.” “And Wilson?” “He went with papa. I don’t think papa could live without Wilson.” “Oh, indeed. I begin to solve the problem of your ghastly little face. You have been housemaid, garde-malade, and bread-winner. Had you no money at all?” Hilda flushed—the quick flush of physical weakness. “Yes, at first,” she replied; “papa gave me quite a lot before going, and that has paid part of the doctor’s bills, and my lessons brought in the usual amount.” “Could you not have given up the lessons for the time being?” “I know you think it dreadful in me to have left mamma for all those afternoons.” Her acceptation of a blame infinitely removed from his thoughts stupefied Odd. “And mamma has thought it heartless, most naturally. But Rosalie is trustworthy and kind. The doctor came three times a day and I can explain to you”—Hilda
- 68. hesitated—“the money papa gave me went almost immediately— some unpaid bills.” “What bills?” Odd spoke sternly. “Why, we owe bills right and left!” said Hilda. “But what bills were these?” “There was the rent of the apartment for one thing; we should have had to go had that not been paid; and then, some tailors, a dressmaker; they threatened to seize the furniture.” “Katherine’s dressmaker?” “Yes; Katherine, I know, never dreamed that she would be so impatient; but I suppose, on hearing that Katherine had gone to England, the woman became frightened.” Peter controlled himself to silence. The very fulness of Hilda’s confidence showed the strain that had been put upon her. “And then,” she went on, as he did not speak, “some of the money had to go to Katherine in England. Poor Kathy! To be pinched like that! She wrote, that at one place it took her last shilling to tip the servants and get her railway ticket to Surrey.” “Why did she not write to me? Considering all things—“ “Oh!” said Hilda—her tone needed no comment—“we have not quite come to that.” She added presently and gently, “I had money for her.” Odd took her hand and kissed it; the glove was loose upon it. “And now,” said Hilda, leaning forward and smiling at him, “you have heard me filer mon chapelet. Tell me what you have been doing.” “My lazy wanderings in the sun would sound too grossly egotistic after your story.” “Has my story sounded so dismal? I have been egotistic, then. I had hoped that perhaps you would write to me,” she added, and a delicately malicious little smile lit her face. Odd looked hard at her, with a half-dreamy stare. “I thought of you,” he said; “I should have liked to write.”
- 69. “Well, in the future do, please, when you feel like it.” Mrs. Archinard was extended on the sofa in the drawing-room when they reached the Rue Pierre Charron. The crisp daintiness of pseudo-invalidism had withered to a look of sickly convalescence. She was much faded, and her little air of melancholy affectation pitifully fretful. “You come before my own daughter, Peter,” she said; “I don’t blame Katherine, since Hilda tells me that she did not let her know of my dangerous condition.” “Not dangerous, mamma,” Hilda said, with a patient firmness not untouched by resentment, a touch to Odd most new and pleasing. “The doctor had perfect confidence in me, and would have told me. I should have sent for papa and Katherine the moment he thought it advisable. Under the circumstances they could have done nothing for you that I did not do.” Hilda had, indeed, rather distorted facts to shield Katherine. What would Mrs. Archinard have said had she known that Katherine, in answer to a letter begging her to return, had replied that she could not? Even in Hilda’s charitable heart that “could not” had rankled. Odd’s despairing gloom discerned something of this truth, as he realized that the uncharacteristic self- justification was prompted by a rebellion against misinterpretation before him. Mrs. Archinard showed some nervous surprise. “Very well, very well, Hilda,” she said, “I am sure I ask no sacrifices on my account. One may die alone as one has lived— alone. My life has trained me in stoicism. You had better wash your face, Hilda. There is a great smudge of charcoal on your cheek,” and, as Hilda turned and walked out, “I have looked on the face of the King of Terrors, Peter. Peter! dear old homely name! the faithful ring in it! It is easy for Hilda to talk! I make no complaint. She has nursed me excellently well—as far as her nursing went. But she has a hard soul! no tenderness! no sympathy! To leave her dying mother every afternoon! To sacrifice me to her painting! At such a time! Ah me!” Large tears rolled down Mrs. Archinard’s cheeks, and her voice trembled with weakness and self-pity. Odd, in his raging resentment, could have exploded the truth upon her; the tears arrested his
- 70. impulse, and he sat moodily gazing at the floor. Mrs. Archinard raised her lace-edged handkerchief and delicately touched away the tears. “I have given my whole life, my whole life, Peter, for my girls! I have borne this long exile from my home for their sakes!” At Allersley Mrs. Archinard had never ceased complaining of her restricted lot, and had characterized her neighbors as “yokels and Philistines.” Speaking with her handkerchief pressed by her finger- tips upon her eyelids, she continued, “I have asked nothing of them but sympathy; that I have craved! And in my hour of need—“ Mrs. Archinard’s point de Venise bosom heaved once more. Odd took her hand with the unwilling yet pitying kindness one would show towards a silly and unpleasant child. “I don’t think you are quite fair,” he said; “Hilda looks as badly as you do. She has had a heavy load to carry.” “I told her again and again to get a garde-malade, two if necessary.” Mrs. Archinard’s voice rose to a higher key. “She has chosen to ruin her appearance by sitting up to all hours of the night, and by working all day in that futile studio.” “Garde-malades are expensive.” Odd could not restrain his voice’s edge. “Expensive! For a dying mother! And with all that is lavished on her studio—canvases, paints, models!” The depths of misconception were too hopelessly great, and, as Mrs. Archinard’s voice had now become shrilly emphatic, he kept silence, his heart shaken with misery and with pity, despairing pity for Hilda. She re-entered presently, wearing on her face too evident signs of contrition. She spoke to her mother in tones of gentle entreaty, humored her sweetly, gayly even, while she made tea. “You know I cannot touch cake, Hilda.” “There are buttered brioches, mamma, piping hot.” “Properly buttered, I hope. Rosalie usually places a great clot in the centre, leaving the edges uneatable.”
- 71. “Mamma is like the princess who felt the pea through all the dozens of mattresses, isn’t she?” said Hilda, smiling at Odd. “But I buttered these with scientific exactitude.” “Exactitude! Ah! the mirage of science! More milk, more milk!” Mrs. Archinard raised herself on one elbow to watch with expectant disapproval the concoction of her tea, and, relapsing on her cushions as the tea was brought to her, “I suppose it is milk, though I prefer cream.” “No, it’s cream.” Hilda should know, as she had herself just darted round the corner to the crêmerie. Odd sprang up to take his cup from her. He thought she looked in danger of falling to the ground. “Do sit down,” he said in a low voice; “you look very, very badly.” “Have you read Meredith’s last?” asked Mrs. Archinard from the sofa. “Hilda is reading it to me in the evenings. We began it, ah! long, long ago. I have sympathy for Meredith, an intimité! It is so I feel, see things—super-subtly. Strange how coarsely objective some minds are! Did you order the oysters for my dinner, Hilda, and the ice from Gagé’s—pistache? I hope you impressed pistache. You will dine with Hilda, of course, Peter; I have my dinner here; I am not yet strong enough to sit through a meal. And then you must talk to me about Meredith. I always find you most suggestive—such new lights on old things. And Verhaeren, too; do you care for Verhaeren? Morbid? Yes, perhaps, but that is a truism—not like you, Peter. ‘Les apparus dans mes chemins,’ poor, modern, broken, bleeding soul! We must talk of Verhaeren. Just now I feel very sleepy. You will excuse me if I simply sans gêne turn over and take a nap? I can often sleep at this hour. Hilda, show Peter the Burne-Jones Chaucer over there. Hilda doesn’t find him limpid, sweet, healthy enough for Chaucer; but nous sommes tous les enfants malades nowadays. There is a beauty, you know, in that. Talk it over.” Hilda and Peter sat down obediently side by side on the distant little canapé before the Burne-Jones Chaucer. They went over the pages, not paying much attention to the woodcuts, but looking down favorite passages together. The description of “my swete” in “The
- 72. Book of the Duchess,” the complaint of poor Troilus, and, once more, Arcite’s death. The quiet room was very quiet, and they looked up from the pages now and then to smile, perhaps a little sadly, at one another. When the dinner was announced Hilda said, as they went into the dining-room— “If your courage fails you, just say so frankly. I have very childish tastes and childish fare.” Indeed, half a cold chicken and a dish of rice constituted the repast. A bottle of claret stood by Odd’s place, and there was a white jar filled with buttercups on the table; but even Rosalie seemed depressed by the air of meagreness, and gave them a rather effaré glance as they sat down. Odd suspected that the cold chicken was in his honor. He had come to the conclusion that Hilda was capable of dining off rice alone. “Delightful!” he said. The chicken and rice were indeed very good, but Hilda saw that he ate very little. “I make no further apologies,” she said, smiling at him over the buttercups; “your hunger be upon your own head.” “I am not hungry, dear.” Hilda had to do most of the talking, but they were both rather silent. It was a happy silence to Hilda, full of a loving trust. When he spoke, it was in a voice of the same gentle fatigue that his eyes showed; but as the eyes rested upon her she felt that the past and the present had surely joined hands.
- 73. O CHAPTER X DD went in the same half-dreamy condition through the morning of the next day. He walked and read, but where he walked and what he read he could hardly have told. He was to fetch Hilda from the Rue d’Assas and go home to tea and dinner with her. His love for Hilda had now reached such solemn heights that his late flight seemed degrading. So loving her, he could not be base. The Rue d’Assas was dreary in a fine drizzling rain. In the Luxembourg Gardens the first young green made a mist upon the trees. It was only half-past four when Odd reached his accustomed post, but hardly had he taken a turn up and down the street when he saw Hilda come quickly from the Lebon abode. She was fully half-an-hour early, but Odd had merely time to note the fact before seeing in a flash that Hilda was in trouble. She looked, she almost ran toward him; and he met her half-way with outstretched hands. “O Peter!” It was the first time she had used his name, and Odd’s heart leaped as her hands caught his with a sort of desperate relief. “Come, come,” she said, taking his arm. “Let us go quickly.” Peter’s heart after its leap began to thump fast. The white distress of her face gave him a dizzy shock of anger. What, who had distressed her? He asked the question as they crossed the road and entered the gardens. Tears now streamed down her face. He had only once before seen Hilda weep, and as she hung shaken with sobs on his arm, the past child, the present Hilda merged into one; his one, his only love. “Let us walk here, dear,” he said; “you will be quieter.” The little path down which they turned was empty, and the fine rain enveloped but hardly wet them. They came to a bench under a
- 74. tree, circled by an unwet area of sanded path. Odd led the weeping girl to it and they sat down. She still held his arm tightly. “Now, what is it?” “O Peter! I can hardly tell you! The brother, the horrible brother.” “Yes?” Peter felt the accumulations of rage that had been gathering for months hurrying forward to spring upon, to pulverize “the brother.” “He made love to me, said awful things!” Odd whitened to the lips. “Tell me all you can.” “I wish I were dead!” sobbed Hilda, “I am so unhappy.” Peter did not trust himself to speak; he took her hand and held it to his lips. “Yes; you care,” said Hilda. She drew herself up and wiped her eyes. “I never thought he would be unpleasant. At times I fancied that he came a good deal into the studio where we worked and, behind his sister’s back, looked silly. But he never really annoyed me. I thought myself unkindly suspicious. To-day Mademoiselle Lebon was called away and he came in. I went on painting. I did not dream —! When, suddenly he put his arms around me—and tried to kiss me!” Hilda gave an hysterical laugh. “Do you know, I had my palette on my hand, and I gave him a great blow with it! You should have seen his head! Oh, to think that I can find that funny now! His ear was covered with cobalt!” Hilda sobbed again, even while she laughed. “He was very angry and horrible. I said I would call his mother and sister if he did not leave me at once, and then—and then”—Hilda dropped her face into her hands—“he jeered at me; ‘You mustn’t play the prude,’ he said.” Odd clenched his teeth. “Hilda, dear,” he said, in a voice cold to severity, “you must go home; I will put you in a cab. I will come to you as soon as I have punished that dog.” “Peter, don’t! I beg of you to come with me. You can do nothing. I must bury it, forget it.” She had risen as he rose.
- 75. “Yes, bury it, forget it, Hilda. He, at least, shall never forget it.” Odd’s fixed look as he led her into the street forced her to helpless silence. “Peter, please!” she breathed, clasping her hands together and gazing at him as he hailed a fiacre. “I will come to you soon. Good-bye.” And so Hilda was driven away. It was past six when Odd reached the Rue Pierre Charron. Rosalie opened the door. Madame was in bed, she had had a bad day. Mademoiselle? she is lying down. She seemed ill. “Et bien malade même,” and had said that she wanted no dinner. “I should like to see her, if only for a moment; she will see me, I think,” said Odd, walking into the drawing-room. Hilda entered almost immediately. She had been crying, and the disorder of her hair suggested that she had cried with her head buried in a pillow, after the stifled feminine fashion. Her face was most pathetically disfigured by tears; the disfigurement almost charming of youth and loveliness; but she looked ill, too. The white cheek and the heavy eyelids, the unsteady sweetness of her lips showed that an extreme of physical exhaustion, as well as the tempest of grief, had swept her beyond all thought of self-control, beyond all wish for it. The afternoon’s unpleasantness had been merely the last straw. The long endurance of the past month—the past months indeed—that had asked no pity, had been hardly conscious of a claim on pity—was transformed by her knowledge of near love and sympathy to a quivering sensibility. There was no reticence in her glance. He was the one she turned to, the one she trusted, the only one who understood and loved her in the whole world. Odd saw all this as the supreme confidence of a supremely reserved nature looked at him from her eyes. He met her, stooping his head to hers, and, like a child, she put up her face to be kissed. When he had kissed her, he drew back. A sudden horrible weakness almost overcame him.
- 76. “Sit down, dear; no, I will walk about a bit. I have been playing the fiery jeune premier to such an extent this afternoon that dramatic restlessness is in keeping.” Hilda smiled faintly, and her eyes followed him as he took a few turns up and down the room. “You look so badly,” he said, pausing before her; “how do you feel?” “Not myself; or, perhaps, too much myself.” Hilda tried to smile, stretching out her arms with a long shaken sigh. “I feel weak and foolish,” she added, clasping her hands on her knee. “It is all right, you know. He apologized profusely.” “How did you make him do that?” “I told him the truth, including the fact of his own despicableness.” “And he believed it?” “I helped him to the belief by a pretty thorough thrashing.” “Oh!” cried Hilda. “He deserved it, dear.” “But—I had exposed myself to it; he thought himself justified.” “I had to disabuse him of that thought. He bawled out something like a challenge under the salutary lesson, but when I promptly seconded the suggestion—insisted on the extreme satisfaction it would give me to have a shot at him—the bourgeois strain came out. He fairly whined. I was disappointed. I had bloodthirsty desires.” “Oh, I am very glad he whined then! Don’t speak of such horrors. You know I am hysterical.” Odd still stood before her, and Hilda put out her hand. “How can I thank you?” He put her hand to his lips, not looking at her but down at the heavy folds of her white dress; it had a shroud- like look that gave him a shudder. Hilda’s life seemed shroud-like, shutting her out from all brightness, from all love—love hers by right, and only hers. “You know, you know that I would do anything for you,” he said.
- 77. The hand he kissed drew him down beside her, hardly consciously, and he yielded to the longing he felt in her for comforting kindness and nearness; yielded, too, to his own growing weakness; but he still held the hand to his lips, not daring to look at her. This childlike trust, this dependence, were dreadful. The long kiss seemed to his troubled soul a momentary shield. He found her eyes on him when he raised his own. “I never thought it would come true—in this way,” she said. “What come true?” “That you would really care for me.” Her pure look seemed to flutter to him, to fold peaceful wings on his breast; its very contentment constituted a caress. The child was still a child, and yet in the look there were worlds of ignorant revelation. A shock of possibilities made Odd dizzy, and the certain strain of weakness in him made it impossible for him to warn and protect her ignorance. He was conscious of a quick grasp at the transcendental friendship of which alone she was aware. “My little friend, I care for you dearly, dearly.” But with the words, his hold on the transcendental friendship slipped, fundamental truths surged up; he took both her hands, and clasping them on his breast, said, hardly conscious of his words— “Sweetest, noblest—dearest,” with an emotion only too contagious, for Hilda’s eyes filled with tears. The sight of these tears, her weakness, the horrible unfairness of her position, appealed, even at this moment, to all his manliness. He controlled himself from taking her into his arms, and his grasp on her hands held her from him. “I understand, Hilda, I understand it all—all you have suffered; the loneliness, the injustice, the dreary drudgery. I know, dear, I know that you have been unhappy.” “Oh yes! I have been unhappy! so unhappy!” The tears rolled down her cheeks while she spoke, fell on Odd’s hands clasping hers. “No one ever cared for me, no one. Papa, mamma, Katherine even,
- 78. not really; isn’t it cruel, cruel?” This self-pity, so uncharacteristic, showing as it did the revulsion in her whole nature, filled Odd with a sort of helpless terror. “That is what I wanted; some one to care; I thought it must be my fault.” The words came in sighing breaths, incoherent: “I have been so lonely.” “My child! My poor, poor child!” “Let me tell you everything. I must tell you now since you care for me. I have been so fond of you—always. You remember when I was a child?” Odd held her hands tightly and mechanically. Poor little hands; they gave him the feeling of light spars clung to in a whirling shipwreck. “Even then I was lonely, I see that now; and even then it weighed upon me, that thought that I was not to the people I loved what they were to me. I felt no injustice. I must be unworthy. It seems to me that all my life I have struggled to make people love me, to make them take me near to them. But you! You were near at once. Do I explain? It sounds morbid, doesn’t it? But it isn’t, for my loneliness was almost unconscious, and I merely felt that with you I was happy, that things were clear, that you understood everything. You did, didn’t you? Only I don’t think you ever quite understood my gratitude, my utter devotion to you.” Hilda’s tears had ceased as she went on speaking, and she smiled now at Odd, a quivering smile. “And then you went away, and I never saw you again. Ah! I can’t tell you what I suffered.” Odd bent his head upon the hands clasped in his. “But how could you have known?” said Hilda tenderly; “I was really very silly and very unreasonable. I thought you would come back because I needed you. I needed the sunshine. Perhaps you were right about the shadow. But for years I waited for you. I felt sure you knew I was waiting. You said you would come back you know; I never forgot that.” She paused a moment: “It all ended in Florence,” she went on sadly; “such a bleak, bitter day, just the day for burying an illusion. I see the cold emptiness of the big room now; oh! the melancholy of it! where I was sitting alone. All came upon me suddenly, the reality. You know those crumbling shocks of
- 79. reality. I realized that I had waited for something that could never come; that you had never really understood, and that it would have been impossible for you to understand. I was a pretty, touching little incident to you, and you were everything to me. I realized, too, how silly it would all seem to any one; how it would be misinterpreted and smiled at as a case of puppy-love perhaps. A sort of cold shame crept through me, and I felt really alone then. Do you know what that feeling is?” Her hand under his forehead lifted his head a little as though to question his face, but putting both her hands over his eyes he would not look at her. “You are so sorry?” Odd nodded. “But you have had that feeling? Imprisoned in oneself; looking, longing for a voice, a smile,—and silence, always, always silence. A thing quite apart from the surface intercourse of everyday life, not touched by it. You have so many friends, so many windows in your prison, you can’t know.” “I know.” “Really?” “Yes, yes.” “And you call out for help and no one hears. Oh, I can’t explain properly; do you understand?” “I understand, dear.” “Well, after that day in Florence, the last cranny of my prison seemed walled up. And—oh, then our troubles came, worse and worse. Responsibilities braced me up—far healthier, of course. And your books! Their strength; their philosophy—don’t tell me I might find it all in Marcus Aurelius; your way of saying it went more deeply in me. Just to do one’s duty; to love people and be sorry for them, and not snivel over oneself. Ah! if you knew all your books had been to me! Would you like it, I wonder?” Again the tenderness, almost playful, in her voice. Odd raised his head and looked at her. “And when I came at last, what did you think?” The loving candor of her eyes dwelt on him.
- 80. “When you came?” she repeated. “Then I saw at once that you were Katherine’s friend, and that your books were the nearest I should ever get to you.” Hilda’s voice hesitated a little; a doubt of the exactitude of her perceptions from this point showed itself in a certain perplexity of tone. “And—I don’t quite understand myself, for I didn’t plan anything—but just because I felt so much I was afraid that you would imagine I made claims on you. I was resolved that you should see that I had reached your standpoint—that I had forgotten—that the present had no connection with the past.” “But I had not forgotten,” Odd groaned. “No?” Hilda smiled rather lightly; “it would have been very strange if you hadn’t. Besides, as I say, I saw at once that you were Katherine’s, and that it was right and natural. Your books taught me, too, the true peace of renunciation, you see! Not that this called for renunciation exactly,” and again Hilda paused with the faint look of perplexity. “There was nothing to renounce since you were hers, except I must have felt a certain disappointment. I felt a little frozen. Such dull egotism!” She turned her eyes away, looking vaguely out into the dusky room. “But even on that first day I meant that you should see, and that she should see, that I knew that the past made no bond: in my heart it might, not in yours, I knew, for all your kindness.” “Go on, Hilda,” said Odd, as she paused. “Well, you know all the rest. When you were engaged and she more than friend, I had hoped for it, and I saw that my turn might come; that I might step into Kathy’s vacated shoes, so to speak; that we might be friends, and all my dreams be fulfilled after all. I began then to let myself know that I did care, for I had tried to help myself before by pretending that I didn’t. I wouldn’t do anything to make you like me. If you were to like me, you would of yourself; all the joy of having you care for me would be in having made no effort. And the dream did come true. I saw more and more that you cared. To- day I feel it, like sunshine.” Odd still stared at her, and again through sudden tears she smiled at him. “Only—isn’t it strange?—things are always so; it must be, too, that I am weak, overwrought, for I feel
- 81. so sad, as though I were at the bottom of the sea, and looking up through it at the sun.” “Great heavens!” muttered Odd. He looked at her for a silent moment, then suddenly putting his arm around her neck, he drew her to him. He did not kiss her, but he said, leaning his head against hers— “And I—so unworthy!” “No, no,” said Hilda, and with a little sigh, “not unworthy, dear Peter.” “I, dully stumbling about your exquisite soul,” Peter went on, pressing her head more closely to his. “Ah, Hilda! Hilda!” “What, dear friend?” “I cannot tell you.” “Unkind; I tell you everything.” “You can tell me everything. You can tell me how much you have cared for me, how much you care. I cannot tell you how much I care. I cannot tell you how infinitely dear you are to me.” He had spoken, her face hidden from him in its nearness; now, turning his head he kissed her hair, and frowning, he looked at her and kissed her on the lips. Hilda drew back and rose to her feet. A subtle change, perplexity deepened, crossed her face, but, standing before him, she looked down at him and he saw that her trust rose as to a test. She put her hands out as though from an impulse to lay them on his shoulders; then, as an instinct within the impulse seemed to warn her, though leaving her clear look untouched, she clasped them together and said gravely— “You may tell me. You are infinitely dear to me.” Odd still frowned. Her terrible innocence gave him a sense of helpless baseness. “I may tell you how much I love you?” and he too rose and stood before her. “I have always loved you,” said Hilda, with her grave look. “I love you now as much as I did when I was a child.”
- 82. The impossible height where she placed him beside her made Odd’s head swim. He felt himself caught up for a moment into the purity of her eyes, and looking into them he came close to her. “My angel! My angel!” he hardly breathed. “Dear Peter,” and the tears came into the pure eyes. And, at the sight, the heaven brimmed with loveliest human weakness, the love unconscious but all revealed, Odd was conscious only of a dizzy descent from impossibility, the crash of the inevitable. One step and he had taken her into his arms, seeing as he did so, in a flash, the white wonder of her face; he could almost have smiled at it—divinely dull creature! Holding her closely, the white folds of the shroud-like dress crushed against his breast, his cheek upon her hair, he could not kiss her and he could not speak, and in a silence as unmistakable as word or kiss, his long embrace forgot the past and defied the future. The painful image of a bird he had once seen, wings broken, dying of a shot and feebly fluttering, came to him as he felt her stir; her hands pushing him away. “Dearest—dearest—dearest.” Her effort faltered to resistless helplessness. Stooping his head he looked at her face; it wore an almost tranquil, a corpse-like look. Her eyes were closed and the eyebrows drawn up a little in a faint, fixed frown; but the childlike line of her mouth had all the sad passivity of death. Odd tremblingly kissed the gentle sternness of the lips. She loved him, but how cruel he was. “Oh, my precious,” he said, “look at me. Forgive me; I love you.” He had freed her hands, and she raised them and bent her face upon them. “You don’t hate me for telling you the truth?” And as she made no sign: “No, no, you don’t hate me; you love me and I love you. I have loved you from the beginning. Oh, my child, my child, why did you let me think you did not care? Look at me, dearest.”
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