Re: [分析] Cardinality、Compact

[jacky7987 (憶)]

因為都獲得大部分的結果所以找了這麼久之前的文章回XD

不過不保證對就是了(￣艸￣)


: 1. Let A be the set of all functions from (0,1) to |R,and
: B be the set of all continuous funtions from (0,1) to |R.
: (a) Show that there is no 1-1 correspondence between |R and A.
: That is,|R and A do not have the same cardinlity.

The cardinal number of S={f|f:X-->Y is a function} is |Y^X|=|Y|^|X|
Hence the cardinal number of A is c^c=2^c>c,where c is the cardinal number
of |R.
雖然這樣好像有點倒過來證明
不過我想這是最快速的想法XDDD

: (b)Prove or disprove that there is a 1-1 correspondence between |R and B.

推文有提到 只要決定再Q(有理數)上的函數值就可以 所以整個的cardinal number
銳減到 N_0^N_0=2^{N_0}=c 因為Q是可數所以跟正整數的cardinal number一樣
所以有bijective的函數


: 2.Let W be a compact subset of |R^n and {Va} be an open cover of W.
: Prove that there is an ε>0 such that for each subset E of W having
: diameter less than ε, there is a V in {Va} containing E.

這個是著名的Lebesgue's Lemma(旁聽高微剛好聽到XD)
所選到的ε稱為Lebesgue number, 題目中用直徑 我證明用半徑 只要再縮小半徑就好了

Proof: Suppose the conclusion is false.

Then for all r>0, there exists x∈W such that

                         B(x;r) is not contained in V_a,

for all a∈I, I is an index set.

Choose r=1/k,k∈|N, and x_k∈W such that

                   B(x_k,1/k) is not contained in any V_a
, a∈I.

Since the conclusion is still fails for W\{x_1},

(If it hold, the we can choose the minimun of radius which is a
contradiction to our hypothesis.)

hence,we may choose x_i≠x_j, for all i≠j.

Hence {x_n} is a infinite set.

Since W is cpt, the all the infinite set has a limit point in W, say x.

Hence x∈V_a, for some a in I, and V_a is open, there exists ε>0 such that

B(x;ε/2)⊆V_a.

Finally, we claim: B(x_k;1/k)⊆B(x;ε)⊆V_a, for some k>>0,

which is a contradiction and we are done.

Since 1/k--->0, choose k>>0 such that 1/k<ε/2, and x_k∈B(x;ε/2).

Given y∈B(x_k;1/k)

|y-x|≦|y-x_k|+|x_k-x|≦1/k+ε/2<ε/2+ε/2=ε

--
※ 發信站: 批踢踢實業坊(ptt.cc) 
◆ From: 111.251.159.236
→ jacky7987   :然後我一直認為2^{N_0}=c是連續統假設( ￣ c￣)y▂ξ
推 plover      :(1) 如何證明 |Y^X|=|Y|^|X| ?
→ dogy007     :2^{N_0}=c 不是連續統假設
→ dogy007     :連續統假設是說N_0和c(N_1)沒有其他 Cardinal number
→ jacky7987   :SOGA
→ jacky7987   :所以有別的方法可以證明他囉?
推 plover      :CH + AC => 2^{N_0}=c
→ jacky7987   :給P大 那好像是定義 Y^X是個符號
→ jacky7987   :等等我再找找QQ
→ jacky7987   :Cardinal exponentiation
→ jacky7987   :|Y^X|=|Y|^|X|
→ jacky7987   :|The set of all function from X to Y|=|Y|^|X|
→ jacky7987   :感謝P大的提示
推 plover      :那為什麼 c^c = 2^c? 看起來應該是 c^c > 2^c
推 plover      :接下來就得證明: c^c > 2^c 以及 2^c >, 然後
→ plover      :再說明 c_1 > c_2 & c_2 > c_3 => c_1 > c_3 ...
→ dogy007     :根據我的理解，N_1 (或者 c) = 2^N_0 可以說是定義
→ dogy007     :我們只是證明了 2^N_0 > N_0 而已
→ dogy007     :而一個要面對的問題就是 N_0, N_1 中間有沒有東西
→ dogy007     :這就是連續統假說
→ dogy007     :先是 Godel 證明了在 ZF 系統下無法否定連續統假說
→ dogy007     :後來 Cohen 又證明了 ZF 系統下無法證明連續統假說
→ dogy007     :於是接不接受連續統假說完全看我們想要怎樣的系統
→ dogy007     :就像原先選擇公設是獨立於其他公設一樣
→ dogy007     :而連續統假說則是即使有了選擇公設也還獨立於系統
→ dogy007     :有些人不想接受選擇公設，但問題是少了選擇公設
→ dogy007     :很多東西都無法證明或者很難證明
→ dogy007     :所以現在絕大多數數學家都接受了選擇公設
→ dogy007     :而連續統假說則非如此
→ keroro321   :c^c=(2^N)^c=2^(Nc)=2^c
→ jacky7987   :感謝樓上幫忙 兩個cardinal number相乘的結果是比較
→ jacky7987   :大的那個
→ jacky7987   :然後cantor說2^c一定嚴格大於c
→ jacky7987   :感謝dogy,其實這篇我只是想要提供點想法而已XD
→ jacky7987   :感謝p大和dogy大這樣熱心:)  幫助我釐清很多事情
→ jacky7987   :也多看了很多文章