There are several ways to show the one-to-one correspondence between P(N) (= the power set of the natural numbers) and the set of the reals (or, equivalently, the set of reals in the interval [0,1).) Continued fractions, and all that. However, one argument is the following (sorry, this argument is from memory, and so I looked for a source; I don't know if the following is a valid source to cite, but if not, please ignore: http://scientiststhesis.tumblr.com/post/52431140481/powersets-infinities-and-the-diagonal-argument): one uses the binary expansion of a real number between 0 and 1 to encode a subset of N (via: construct set S such that: if nth place in expansion =0, then n not in S; nth place in expansion =1, then n is in S.) However, this encoding actually only makes a correspondence between ordered subsets of N and [0,1). Since there are many elements in the set of ordered subsets of N for every element in P(N), this only proves that |P(N)| ≤ |R|. So, how does one then make the correspondence between the set of ordered subsets of N and P(N)?(adsbygoogle = window.adsbygoogle || []).push({});

(The source cited above starts with the classical diagonal argument about |N|≤|R| before stating the above attempt at |P(N)| = |R|.)

Thanks for any feedback.

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# Missing step in standard proof of |P(N)|=|R|

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