Cantor-Bernstein-Schroeder Theorem

3月30日 01:34
小弟想問個關於此定理的證明問題,先給出我當初讀的證明版本: 定理:
證明:
證明我大致上看完了,但小弟有幾個問題想請教,就是這個證明是不是已經假定了A,B兩個集合都要是無窮集合(Infinite set)的情況呢? 因為如果要一直取A0,A1,A2.....這樣在A中的子集合下去,那麼A,B的元素個數應該是要有無窮多個吧? 第二個問題是,此證明是不斷的在A和B的子集之間構造1-1函數, 然後再考慮這樣一個映射h,且驗證它是不是1-1且onto函數嗎? 不知道有沒有理解錯.... 還有一個問題,A0和A1中的元素能有交集嗎? 我看證明中的圖示好像是都沒有交集。 比如A0={1,2,3}到f(A0)=f{1,2,3,}={4,5,6} 再由g得g(f{4,5,6})=7,8,9? 我只知道A0和A1的元素個數一樣,但元素能不能有交集還請大佬解答🙏 謝謝各位🙏🙏🙏
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國立臺灣大學 數學系
因為用中文很難敘述,所以才用英文。 (0) Let's check a detail before answering your questions. In the proof, we have a function h(x). You should check that this function is well-defined. What does this mean? If x is in A_{\infty}, then h(x):=f(x) has no problem. If x is not in A_{\infty}, then h(x):=g^{-1}(x). In this case, we have to be careful because g^{-1}(x) may be empty. Therefore, to check that h(x) is well defined is equivalent to check that "x is not in A_{\infty} => x is in im(g)". Make sure you can show this. Now we come back to your questions. (1) Can |A| be finite? It is not important whether |A| is infinite or not. If |A| is finite, then A_m will be an empty set for m large enough. If it is the case, we still can take infinite union of all A_m. (2) The idea of the proof Your understanding of the proof is a little bit rough. The spirit of the proof is to construct a bijective function from A to B. As you know, if f:A->B is an injection, then f:A->im(f) is a bijection. So the task is to "enlarge" im(f) such that im(f)=B. The method we use here is to construct a sequence of subsets A_n such that (*) A\im(g) is contained in A_{\infty}. The property (*) is actually a reformulation of "enlarging im(f)". (3) The intersection of A_0 and A_1 Yes, the intersection of A_0 and A_1 is empty. To see this, take any x in A_1. By definition of A_1, we can write x=g(f(x_0)) for some x_0 in A_0. Since f(x_0) is in B, the element x is particularly in g(B). However, A_0=A\g(B). Thus we show that x is not in A_0. That is, the intersection of A_0 and A_1 is empty.
B1 感謝你🙏 其實證明中有說到為什麼h是well-defined,但整個過程太長我就簡短的放上來。 問題(2)還是不太懂.... 想問對於證明的idea那裡,(*)當中的X\im(g)的X是指什麼集合?
國立臺灣大學 數學系
B2 抱歉我打錯了,是A不是X
B3 我想再check一下證明的idea,因為我還不太了解..... 我們的目的是要找到從A到B的1-1且onto,f是從A到B的1-1,那麼f從A到A的image是1-1且onto,然後我們要讓f從A到A的image擴大到整個B上,使得f從A到B是1-1且onto,這個我能理解。 但後面我就不太懂了,能再解釋一下嗎? 感謝🙏
國立臺灣大學
B5 所以證明中之所以會將h定義成:
是為了不讓f(A0)同時被它自己所形成的image打到,且又同時被g(B)的inverse function打到,造成f(A0)這塊不是bijection的情況嗎? 總上述,才會有下圖這個映射構造的嘛?
國立臺灣大學 數學系
B6
B7 那我清楚了 感謝你🙏