Tilted Forum Project Discussion Community - View Single Post - Cantor-Schroder-Bernstein Theorem

dimbulb · 09-19-2003, 06:07 AM

Well, there exists a bijection between S and a subset of T, T1(call the f), and a bijection between T and a subset of S,S1 (call it g).

Restrict g, so that its domain is f(S). so there is a bijection between f(S) and g(f(S)). lets call the set g(f(S))

observe that g(f(S)) is a proper subset of S1, so lets call g(f(S)) S2.

so we have a bijection between S and f(S), and a bijection between f(S) and S2. Thus S ~S2.
S2 is a proper subset of S1 which is a proper subset of S. So if you can prove that S~S1, and since there is a bijection between S1 and T, you can obtain a bijection between S and T.

I've solved the problem without using my reformulation, but using the above principle of obtaining bijections between a set and its proper subset.

took me a while, and finally got it. It occured to me while i was reading a proof for my reformulation..heh...

09-19-2003, 06:07 AM	#13 (permalink)
dimbulb Riiiiight........	Well, there exists a bijection between S and a subset of T, T1(call the f), and a bijection between T and a subset of S,S1 (call it g). Restrict g, so that its domain is f(S). so there is a bijection between f(S) and g(f(S)). lets call the set g(f(S)) observe that g(f(S)) is a proper subset of S1, so lets call g(f(S)) S2. so we have a bijection between S and f(S), and a bijection between f(S) and S2. Thus S ~S2. S2 is a proper subset of S1 which is a proper subset of S. So if you can prove that S~S1, and since there is a bijection between S1 and T, you can obtain a bijection between S and T. I've solved the problem without using my reformulation, but using the above principle of obtaining bijections between a set and its proper subset. took me a while, and finally got it. It occured to me while i was reading a proof for my reformulation..heh...