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

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