5.1 Defining and visualizing functions. Dynamic slides.

Introduction to set theory and to methodology and philosophy of
mathematics and computer programming
Deﬁning and visualizing functions
An overview
by Jan Plaza
c 2017 Jan Plaza
Use under the Creative Commons Attribution 4.0 International License
Version of November 4, 2017

Existence and uniqueness
How to express in ﬁrst-order logic “there exists unique y s.t. A(y)”
without using the abbreviation ∃!y A(y) ?

Existence and uniqueness
Existence: ∃y A(y).
Uniqueness: ∀y1,y2 (A(y1) ∧ A(y2) → y1 =y2).

Existence and uniqueness; Functions
Deﬁnition
A function is any binary relation f such that ∀x ∀y1,y2 (xfy1 ∧ xfy2 → y1 =y2),
or equivalently, ∀x (∃y xfy → ∀y1,y2 (xfy1 ∧ xfy2 → y1 =y2)).

Deﬁnition
This means: if there is y s.t. xfy then such a y is unique.

Deﬁnition
This means: if there is y s.t. xfy then such a y is unique.
Deﬁnition. Let f be a function.
1. Let x ∈ domain(f). The value of f at x , f(x) , is the unique y such that xfy.
2. f maps x to y , denoted f : x → y or x
f
→ y , if xfy.

x
Is this a function? No. More than one value of y is paired with the same value of x.

Are these functions?
x
No.
x
Yes.

{ −1, 1 , 0, 0 , 1, 1 },
i.e.
x y
−1 1
0 0
1 1
{ 1, 0 , 0, 1 , 1, 2 },
i.e.
x y
1 0
0 1
1 2

{ −1, 1 , 0, 0 , 1, 1 },
i.e.
x y
−1 1
0 0
1 1
Yes.
{ 1, 0 , 0, 1 , 1, 2 },
i.e.
x y
1 0
0 1
1 2
No.

-1
0
1
-1 0 1
x
0
1
2
0 1
x

-1
0
1
-1 0 1
x
Yes.
0
1
2
0 1
x
No.

-1
0
1
0
1
0
1
0
1
2

-1
0
1
0
1
Yes.
0
1
0
1
2
No.

0
-1 1
0
1 2

0
-1 1
Yes.
0
1 2
No.

Terminology: total/partial, on
f is on X or f is a (total) function on X if domain(f) = X.
Definition. f is a partial function on X if there is X ⊆X s.t. f is a function onX .
Fact. If f is a partial function on X, then f is a total function on domain(f).
If f is a function and X ⊇ domain(f), then f is a partial function on X.
Fact. Consider these conditions:
1. f is a binary relation,
2. domain(f) ⊆ X,
3. for every x ∈ X, f maps x to at most one value,
4. for every x ∈ X, f maps x to least one value.
Then:
f is a function on X iff f satisfies conditions 1-4;
f is a partial function on X iff f satisfies conditions 1-3.

Vertical line tests on the Cartesian plane are the following.
Let G be a subset of the Cartesian plane.
G is the graph of a function from R to R
iff every straight line parallel to the y axis intersects G in exactly one point.
G is the graph of a partial function from R to R
iff every straight line parallel to the y axis intersects G in at most one point.
Vertical line tests in discrete Cartesian diagrams are the following.
Let G be a subset of X × Y in a discrete Cartesian diagram.
G is the graph of a function from X to Y
iff every column in the diagram contains exactly one point of G.
G is the graph of a partial function from X to Y
iff every column in the diagram contains at most one point of G.

Terminology: from, to
1. f is to/into Y if range(f) ⊆ Y .
2. f is (a total function) from X to Y , denoted f : X −→ Y or X
f
−→ Y ,
if domain(f)=X and range(f) ⊆ Y .
Definition
f is a partial function from X to Y , denoted f : X −→ Y or X
f
−→ Y ,
if there exists X ⊆ X such that f : X −→ Y .
f : X −→ Y f : X −→ Y
domain(f) ⊆ X domain(f) = X
range(f) ⊆ Y range(f) ⊆ Y
Our definitions imply that the term “to/into Y ” applies to partial functions as well:
for any partial function f, f is to/into Y iff range(f) ⊆ Y .

Exercise
Is this a function on R?

Exercise
Is this a function on R? No.

Exercise
Is this a partial function on R?

Exercise
Is this a partial function on R? Yes.

Exercise
What is the domain of this function?

Exercise
What is the domain of this function? R − {0}.

Exercise
Is this a function on R − {0}?

Exercise
Is this a function on R − {0}? Yes.

Exercise
Is it correct to say f : R −→ R?

Exercise
Is it correct to say f : R −→ R? No.

Exercise
Is it correct to say f : R − {0} −→ R?

Exercise
Is it correct to say f : R − {0} −→ R? Yes.

Exercise
Is it correct to say f : R −→ R?

Exercise
Is it correct to say f : R −→ R? Yes.

Exercise
Is it correct to say f : R − {0} −→ R?

Exercise

Example
1. Let f ={ 0, −1 , 1, 0 , 2, 3 }.
f is a function from {0, 1, 2} to Z.
f is also a function from {0, 1, 2} to {−1, 0, 3}.
2. Alternatively we could specify it as:
a function f on {0, 1, 2} s.t. f(0)=−1, f(1)=0, f(2)=3.
a function f on {0, 1, 2} s.t. f : 0 → −1, f : 1 → 0, f : 2 → 3.
a function f on {0, 1, 2} s.t. f(x)=x2 − 1.
5. Let g be a function on {0, 1, 2, 3} s.t. g(x)=x2 − 1.
Although g and f are defined by the same formula,
they are different functions
because they have different domains.

Example
1. Expression y =
√
1 − x2 does not define a function on R,
but it defines a function on [−1, 1].
2. Expression y = ±
√
1 − x2 specifies coordinates of points of a unit circle,
however it does not define a function on [−1, 1].

Exercise
Consider the formula y =
√
x.
(Recall that for x = 9 we have just y = 3,
because the deﬁnition of square root requires it to be non-negative.)
1. Does this formula deﬁne a function from R to R?

Exercise
√
x.
1. Does this formula deﬁne a function from R to R? No.

Exercise
√
x.
2. Does this formula deﬁne a function from {u ∈ R : u 0} to R?

Exercise
√
x.
2. Does this formula deﬁne a function from {u ∈ R : u 0} to R? Yes.

Exercise
√
x.
3. Does this formula deﬁne a partial function from R to R?

Exercise
√
x.
3. Does this formula deﬁne a partial function from R to R? Yes.

Exercise
√
x.
4. Does this formula deﬁne a function from {u ∈ R : u 0} to Q?

Exercise
√
x.
4. Does this formula deﬁne a function from {u ∈ R : u 0} to Q? No.

Exercise
√
x.
5. Does this formula deﬁne a function from {u ∈ R : u 0} to {u ∈ R : u 0}?

Exercise
√
x.
5. Does this formula deﬁne a function from {u ∈ R : u 0} to {u ∈ R : u 0}?
Yes.

Exercise
Are the following statements correct?
1. Let f1, f2 be functions.
If domain(f1) = domain(f2) and f1 ⊆ f2, then f1 = f2.

Exercise
If domain(f1) = domain(f2) and f1 ⊆ f2, then f1 = f2. Yes.

Exercise
2. Let R1, R2 be binary relations.
If domain(R1) = domain(R2) and R1 ⊆ R2, then R1 = R2.

Exercise
If domain(R1) = domain(R2) and R1 ⊆ R2, then R1 = R2. No.

Exercise
3. Let f1, f2 be partial functions on the same set X.
If f1 ⊆ f2, then f1 = f2.

Exercise
3. Let f1, f2 be partial functions on the same set X.
If f1 ⊆ f2, then f1 = f2. No.
Whenever the answer is negative, provide a counter-example.

Note
1. An attempt to define f : Q −→ Z:
f(m
n ) = m + n, where m, n ∈ Z.
This is not correct, because
1
2 = 2
4 but f(1
2) = 1 + 2 = 3=6 = 2 + 4 = f(2
4)
violating ∀x,y1,y2 (xfy1 ∧ xfy2 → y1 =y2).
2. An attempt to define g : Q −→ Z:
let g(0)=1 and
let g(m
n ) = m + n where m, n ∈ Z, m=0, n > 0,
and m and n do not have a common divisor greater than 1.
This is correct.
(We have used a canonical form of rational numbers.)
3. Every rational number has many representations (1
2 = 2
4 = ...).
To be correct, the definition must be
independent of the representation .

Exercise
Let Q+ be the set of all positive rational numbers.
Are these attempts to deﬁne a function correct?
1. f : Q −→ Q s.t. f(m
n ) = n
m where m, n ∈ Z.

Exercise
1. f : Q −→ Q s.t. f(m
n ) = n
m where m, n ∈ Z. No.

Exercise
1. f : Q −→ Q s.t. f(m
n ) = n
2. f : Q+ −→ Q+ s.t. f(m
n ) = n
m where m, n ∈ Z.

Exercise
1. f : Q −→ Q s.t. f(m
n ) = n
2. f : Q+ −→ Q+ s.t. f(m
n ) = n
m where m, n ∈ Z. Yes.
Show this by proving:
if m1, n1, m2, n2 ∈ Z and m1
n1
, m2
n2
∈ Q+ and m1
n1
= m2
n2
then n1
m1
= n2
m2
.

Note. An attempt to to model the concept “mother of”:
1. Persons is a non-empty set,
2. Persons is a ﬁnite set,
3. motherOf : Persons −→ Persons
(i.e. every person has a unique mother, who is a person),
4. motherOf is a function such that for every p in its domain:
p=motherOf(p),
p=motherOf(motherOf(p)),
p=motherOf(motherOf(motherOf(p))),
etc. (i.e. there are no cycles).
These conditions are contradictory.
(The same problem occurs with theSupervisorOf in employee databases.)
To see why, try drawing directed graphs
showing a function motherOf on a three-element set Persons.

5.1 Defining and visualizing functions. Dynamic slides.

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5.1 Defining and visualizing functions. Dynamic slides.