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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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marked as duplicate by Asaf Karagila, Brian M. Scott, Willie Wong Jul 5 '12 at 8:43

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It ways that there is a bijection between the real numbers and the collection of all subsets of $\mathbb{N}$. There is a way of labeling each subset of $\mathbb{N}$ with exactly one real number, in such a way that each subset corresponds to a unique real number, each real number corresponds to a subset, and different real numbers correspond to different subsets. –  Arturo Magidin Jul 4 '12 at 17:46
    
@ArturoMagidin that is what I meant. You are correct about 2^n. –  Inquest Jul 4 '12 at 17:49
    
@Arturo Magidin - the cardinality is $\aleph_0$ not $\mathbb{N}$. you wrote the set of all functions from $\mathbb{N}$ to the set with two elements that are the empty set and the set that have one element that is the empty set. –  Belgi Jul 4 '12 at 17:50
    
@Belgi: Yes, I forgot the bars (I didn't want to use alephs, because the OP wasn't using them). –  Arturo Magidin Jul 4 '12 at 17:52
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Relevant post: The simplest way of proving that |P(N)| = |R| –  dtldarek Jul 4 '12 at 17:57

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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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Note that there are more than one way to represent the same number so this isn't exactly what the result is. but since there are only countable repetitions the result is the same –  Belgi Jul 4 '12 at 17:56
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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. –  tomasz Jul 4 '12 at 17:59

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