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The simplest way of proving that $|\mathcal{P}(\mathbb{N})| = |\mathbb{R}| = c$

I was reading Rubin and came across the fact that $2^{\aleph_0}$ is the cardinality of reals.

I follow the proof but I cant seem to understand physically what this attempts to say. Is there a nice intuitive explanation for this property?

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    @Inquest Great I was able to help. Please remember to upvote the answers you like for the future readers.2012-07-04

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$2$ raised to the cardinality of the natural numbers, means all functions from the natural numbers to a set of two members (lets say $\{0,1\}$) which practically means all sequences composed of $0$'s and $1$'s. Think of every number in its binary representation. If the number is rational, the sequence will just have an infinite number of zeros from a certain point (we'll avoid representations of numbers similar to $0.999\ldots$ in decimal base). If the number is irrational, it will be any (non constant from a certain point) infinite sequence of zeros and ones. So, any real number can be represented as an infinite sequence of zeros and ones, and that's exactly what the above equality of cardinals asserts.

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    The number will only have only zeroes after a certain point iff it is of the form $n\cdot 2^{-k}$. For example, one third will be $\frac{1}{11}=0,(01)$ with a period. Irrational numbers will be those without periodic expansions.2012-07-04