Proving the Equivalence of Cardinalities |P(N)| = |Reals|

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Homework Statement



Show that |P(N)|=|R|. (R=reals, |X| is the cardinality of X).

Homework Equations





The Attempt at a Solution



#1.) P(N) --> R

Given any element, A, of P(N), construct a decimal expansion .x1x2x3x4... by the rule that x_i=1 (if i is in A) and x_i=0 (if i is not in A).

So the element {1,7} would give .1000001

This map is 1-1 but not onto the Reals.

#2.) R --> P(N)

If I can show this direction then #3 follows. I know a little about the mapping of [0,1] onto the Reals. I cannot determine if that helps here.

#3.) Using the Cantor-Schroeder-Bernstein Theorem, |P(N)|=|R|.
 
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For #1, instead of a decimal expansion make it a binary expansion. In fact I think you can show this to give a bijection between [0, 1] and P(N). Then use (or prove) the "famous" result that | [0, 1] | = |R|.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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