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Manifold
September 18th, 2004, 5:03 AM
I've got an exercise I would like to discuss with you...I came to this idea because of the thread "even and odd numbers" which has a lot to do with it...

Task (Source: V.A. Zorich, Mathematical Analysis 1, Springer-Verlag):

a) Prove the equipollence of the closed interval \{x\in\mathbb{R}~|~0\le{x}\le{1}\} and the open interval \{x\in\mathbb{R}~|~0<x<1\} of the real line \mathbb{R} both using the Schröder-Bernstein theorem and by direct exhibition of a suitable bijection.

b) Analyze the following proof of the Schröder-Bernstein theorem:
(card~X\le{card~Y})\wedge(card~Y\le{card~X}) \Rightarrow (card~X=card~Y).
Proof:
It suffices to prove that if the sets X,Y,Z are such that X\supset{Y}\supset{Z} and card~X=card~Z, then card~X=card~Y. Let f:X\rightarrow{Z} be a bijection. A bijection g:X\rightarrow{Y}can be defined, for example, as follows:
g(x)=\left\{{f(x),~if~x\in{f^n(X)\setminus{f^n(Y)} ~for~some~n\in\mathbb{N},}\atop~{x,~otherwise.}\
Here f^n=f\circ{...}\circ{f} is the nth iteration of the mapping f and \mathbb{N} is the set of natural numbers. (Remark: N={1,2,3,...} in this terminology)

matt grime
September 21st, 2004, 11:37 AM
what did you want to discuss about them? the first is easy by S-B, and the direct proof is the standard: if X is an infinite set there is an countable (infinite) subset, let

x_1=0 , x_2=1/2 x_3=1/3, x_4 =1/4 etc and define bijection from [0,1] to (0,1] be

y to y if y not one of the x_i

x_i to x_{i+1}

other wise.

clearly a bijection. it is then trivial to show (0,1] ~ (0,1) by a similare argument.

the second question is a fancy proof of S-B, but no different from the usual one really.