Lect2 230708501

26
A 2008 A

description

math

Transcript of Lect2 230708501

Page 1: Lect2 230708501

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Page 2: Lect2 230708501

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�êÆ�Ù§©|Ø��Ó§8ÜØ´��<Õg��z"ù�<Ò´x÷"8ÜØ´ïÄ“Ã�"Vg�êÆ"

Georg Cantor:1845-1918

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Page 3: Lect2 230708501

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Page 4: Lect2 230708501

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I protest above all against the use of an infinite quantity asa completed one§which in mathematics is never allowed.The infinite is only a manner of speaking§in which oneproperly speaks of limits.

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Page 5: Lect2 230708501

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Page 6: Lect2 230708501

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Page 7: Lect2 230708501

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CBS Theorem��{

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Page 8: Lect2 230708501

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Page 9: Lect2 230708501

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CBS Theorem��{

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Page 10: Lect2 230708501

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CBS Theorem��{

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Page 11: Lect2 230708501

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Page 12: Lect2 230708501

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CBS Theorem��{

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Page 13: Lect2 230708501

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CBS Theorem��{

Cantor-Bernstein-Shröder Theorem

½n (CANTOR-BERNSTEIN-SCHRÖDER THEOREM)

�A, B´ü�8ܧf : A → B Ú g : B → A´ü�N�"XJfÚgÑ´ü�§KkV�h : A → B.

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Page 14: Lect2 230708501

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CBS Theorem��{

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Proof.PC = g[B], h = g ◦ f . 8B½ÂAi , Ci , DiXeµ

A0 = A, C0 = C, D0 = A0 \ C0;An+1 = h[An], Cn+1 = h[Cn], Dn+1 = An+1 \ Cn+1.

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Page 15: Lect2 230708501

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CBS Theorem��{

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Proof.

-D∗ = A \⋃∞

i=0 Di . �f , gÑØ´÷�§Kk±eäóµé?¿i ≥ 0, Ai+1 ⊂ Ci ⊂ Ai ;D0, D1, · · · , ...üüØ��¶� h(Di) = Di+1.A =

⋃∞i=0 Di ∪ D∗;

C =⋃∞

i=1 Di ∪ D∗.

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Page 16: Lect2 230708501

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CBS Theorem��{

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Proof.½Â¼êk : A → CXeµ

f (a) =

{h(a), a ∈

⋃∞i=0 Di

a, otherwise.

Kk´V�§l A�C�³§dC�B�³§·��A�B��³"y²�."

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Page 17: Lect2 230708501

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CBS Theorem��{

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Page 18: Lect2 230708501

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CBS Theorem��{

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Page 19: Lect2 230708501

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CBS Theorem��{

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A = {♣,♦,♥,♠}

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♣ ♦ ♥ ♠l l l l

{♦,♥} {♦,♠} {♣,♦,♥} {♣,♥,♠}

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Page 20: Lect2 230708501

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♣ ♦ ♥ ♠♣ − + +♦ + ⊕ + −♥ + − ⊕ +♠ − + − ⊕

♣ ♦ ♥ ♠l l l l

{♦,♥} {♦,♠} {♣,♦,♥} {♣,♥,♠}

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Page 21: Lect2 230708501

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~X§éþ��Ñ�o�s@§·�kT = {♣}.ù`²¹o����8ÜAÚ§��8�mØUïá��éA"

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Page 22: Lect2 230708501

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1 2 3 4 · · ·l l l l · · ·

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M = {i : i 6∈ Mi}.

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Page 23: Lect2 230708501

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Page 24: Lect2 230708501

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þ¡�(ii)��±��e¡�(ii′) é?¿n§XJ P(0), P(1), · · · , P(n)Ѥá§KP(n + 1)�¤á"

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Page 25: Lect2 230708501

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“é?¿��ên, n2 + 5n + 1Ñ´óê""PP�5�“n2 + 5n + 1´óê""b�P(n)ý§·�y²P(n + 1)�ý"

Proof.

b�P(n)�ý§Kn2 + 5n + 1´óê"d(n + 1)2 + 5(n + 1) + 1 = (n2 + 5n + 1) + 2(n + 3)§·���(n + 1)2 + 5(n + 1) + 1�´óê§=P(n + 1)�ý"

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Page 26: Lect2 230708501

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