Manifold Posted September 18, 2004 Share Posted September 18, 2004 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 [math]\{x\in\mathbb{R}~|~0\le{x}\le{1}\}[/math] and the open interval [math]\{x\in\mathbb{R}~|~0<x<1\}[/math] of the real line [math]\mathbb{R}[/math] both using the Schröder-Bernstein theorem and by direct exhibition of a suitable bijection. b) Analyze the following proof of the Schröder-Bernstein theorem: [math](card~X\le{card~Y})\wedge(card~Y\le{card~X}) \Rightarrow (card~X=card~Y)[/math]. Proof: It suffices to prove that if the sets [math]X,Y,Z[/math] are such that [math]X\supset{Y}\supset{Z}[/math] and [math]card~X=card~Z[/math], then [math]card~X=card~Y[/math]. Let [math]f:X\rightarrow{Z}[/math] be a bijection. A bijection [math]g:X\rightarrow{Y}[/math]can be defined, for example, as follows: [math]g(x)=\left\{{f(x),~if~x\in{f^n(X)\setminus{f^n(Y)}~for~some~n\in\mathbb{N},}\atop~{x,~otherwise.}\[/math] Here [math]f^n=f\circ{...}\circ{f}[/math] is the nth iteration of the mapping [math]f[/math] and [math]\mathbb{N}[/math] is the set of natural numbers. (Remark: N={1,2,3,...} in this terminology) Link to comment Share on other sites More sharing options...
matt grime Posted September 21, 2004 Share Posted September 21, 2004 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. Link to comment Share on other sites More sharing options...
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