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