Talk:Cardinality of Power Set of Natural Numbers Equals Cardinality of Real Numbers

Yeeaahhhh.....wouldn't this basically prove the Continuum hypothesis? EmperorZelos (talk) 19:06, 15 May 2015 (UTC)
 * No, the Continuum Hypothesis is that there is no set with cardinality strictly less than the cardinality of the real numbers and strictly greater than the cardinality of the natural numbers. The named theorem is far more elementary, though the proof is still rather difficult to do rigorously. I shall write a proof up over summer after my exams have finished, unless someone beats me to it. Oliver (talk)

$|\R| = 2^{\aleph_0}$ always holds. The continuum hypothesis is the assertion $2^{\aleph_0} = \aleph_1$. &mdash; Lord_Farin (talk) 20:36, 15 May 2015 (UTC)

Fair enough EmperorZelos (talk) 02:46, 16 May 2015 (UTC)
 * Altough wouldn't it be fairly trivial? Considering $2^\N$ contains all subsets of $\N$ including infinite sets, combined with binary representation of real numbers, one can easily construct a bijection between $[0,1]$ where if the n-th decimal is 1, then $n\in f(a)$ and if it's 0 it is not, or? EmperorZelos (talk) 07:44, 16 May 2015 (UTC)


 * It is, except for the fact that you have to be careful that some rational numbers have two binary representations. I'm pretty confident there are good proofs over at Math.StackExchange. &mdash; Lord_Farin (talk) 08:37, 16 May 2015 (UTC)
 * Naturally but that's just a minor inconvinience EmperorZelos (talk) 08:44, 16 May 2015 (UTC)