Infinite Sets and the Schröder-Bernstein Theorem

[Manifold]

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)

[matt grime]

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.