![CBS Theorem
CBS
Proof.
C = g[B], h = g ◦ f . Ai , Ci , Di
A0 = A, C0 = C, D0 = A0 C0 ;
An+1 = h[An ], Cn+1 = h[Cn ], Dn+1 = An+1 Cn+1 .
A](https://image.slidesharecdn.com/lect2230708501-090725095917-phpapp01/85/Lect2-230708501-14-320.jpg)
Lect2 230708501
- 1. A 2008 A
- 2. “ " Georg Cantor: 1845-1918 A
- 3. Cantor 2 A
- 4. Cantor ( ) (1777-1855) I protest above all against the use of an infinite quantity as a completed one which in mathematics is never allowed. The infinite is only a manner of speaking in which one properly speaks of limits. A
- 5. Cantor (1873) (1873) A
- 6. 1873 1874 1891 A
- 7. CBS Theorem Outline 1 CBS Theorem 2 A
- 8. CBS Theorem A
- 9. CBS Theorem ( ) ( ) A B A ≈ B, A B A
- 10. CBS Theorem ( ) ( ) A ( ) |A|, A |A| = |B| A ≈ B. A
- 11. CBS Theorem ( ) A |A| = |N|. A A A
- 12. CBS Theorem Outline 1 CBS Theorem 2 A
- 13. CBS Theorem Cantor-Bernstein-Shröder Theorem (C ANTOR -B ERNSTEIN -S CHRÖDER T HEOREM) A, B f :A→B g:B→A f g h : A → B. A
- 14. CBS Theorem CBS Proof. C = g[B], h = g ◦ f . Ai , Ci , Di A0 = A, C0 = C, D0 = A0 C0 ; An+1 = h[An ], Cn+1 = h[Cn ], Dn+1 = An+1 Cn+1 . A
- 15. CBS Theorem CBS Proof. ∞ D∗ = A i=0 Di . f,g i ≥ 0, Ai+1 ⊂ Ci ⊂ Ai ; D0 , D1 , · · · , ... h(Di ) = Di+1 . ∞ ∗ A= i=0 Di ∪ D ; ∞ ∗ C= i=1 Di ∪ D . A
- 16. CBS Theorem CBS Proof. k :A→C h(a), a ∈ ∞ Di i=0 f (a) = a, otherwise. k A C C B A B A
- 17. CBS Theorem Outline 1 CBS Theorem 2 A
- 18. CBS Theorem (1891) A A A A
- 19. CBS Theorem A = {♣, ♦, ♥, ♠} A ♣ ♦ ♥ ♠ {♦, ♥} {♦, ♠} {♣, ♦, ♥} {♣, ♥, ♠} A
- 20. CBS Theorem + −. ♣ ♦ ♥ ♠ ♣ − + + ♣ ♦ ♥ ♠ ♦ + ⊕ + − ♥ + − ⊕ + {♦, ♥} {♦, ♠} {♣, ♦, ♥} {♣, ♥, ♠} ♠ − + − ⊕ A
- 21. CBS Theorem 4 T: T T = {♣}. A A
- 22. CBS Theorem f N 1 2 3 4 ··· ··· M1 M2 M3 M4 ··· f f i Mj (i, j) − +. M = {i : i ∈ Mi }. M = Mn n f A
- 23. P P n P(n) P (i) ( ) P(0) (ii) ( ) n P(n) P(n + 1) 1 1 P(n) A
- 24. (ii) (ii ) n P(0), P(1), · · · , P(n) P(n + 1) (1)+(ii ) P P (ii ) n P(m) m<n P(n) A
- 25. “ n, n2 + 5n + 1 " P “n2 + 5n + 1 " P(n) P(n + 1) Proof. P(n) n2 + 5n + 1 (n + 1)2 + 5(n + 1) + 1 = (n2 + 5n + 1) + 2(n + 3) (n + 1)2 + 5(n + 1) + 1 P(n + 1) A
- 26. P, A