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Is $2^{|\mathbb{N}|} = |\mathbb{R}|$? If so, how?

I was reading the Wiki page on the , and it says "Moreover, $\mathbb{R}$ has the same number of elements as the power set of $\mathbb{N}$", but I don't see how this is true?

I feel like it has something to do with binary, but I'm not too sure how it works? Do I have to show a map of all reals can be done in binary? I'm just very confused, and any advice would be appreciated!

Martin Sleziak
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Charles D
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  • What on earth does "a map of all reals can be done in binary" mean? – Chris Eagle Oct 08 '12 at 18:55
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    Have you tried searching this site for the answer? – Asaf Karagila Oct 08 '12 at 19:06
  • See also: http://math.stackexchange.com/questions/553526/the-set-of-real-numbers – Martin Sleziak Aug 09 '16 at 05:16
  • and see this http://math.stackexchange.com/questions/1885700/cardinality-of-power-set-of-mathbb-n-is-equal-to-cardinality-of-mathbb-r/1885782 . Basically you can map ${0,1}^N$ to [0,1] via mapping $(x_1,x_2, x_3,..... )$ to $\sum x_i/2^i$. The sum is a decimal expansion but written in binary. So the set of such sums is all reals between [0.1]. It's a simple matter to show $|[0,1]| = |\mathbb R|$. So we have $|{0,1}^N| = 2^N = |[0,1]| = |\mathbb R|$ – fleablood Aug 09 '16 at 05:34

2 Answers2

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In short: A binary number $0.a_1a_2a_3\ldots$ can be identified with the set $\{n\in \mathbb N\mid a_n\ne 0\}$. A few details have to be checked, though

Hagen von Eitzen
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  • Amplifying on this, we have issues like the fact that 1.00000... and 0.11111... represent the same number. These issues don't turn out to have any effect on the result of the argument, because there are in some sense very few of these examples -- most real numbers do not end in 000... or 111... –  Oct 08 '12 at 20:03
  • This maps the set of all {a1,a2,a3...} to {0.a1a2a3....} = [0.0000....,0.111111....] = [0,1] without duplicates as 0.11111.... = 1 but we don't have 1.0 represented. So that's fine. But we've shown it is 1-1 to [0,1] and not the reals. We can map [0, \infty) to [0,1) by mapping x to x/(x+1) and we can map (-\infty, 0] to (- 1, 0] by mapping x to x/(x - 1) so we can map (-infty, infty) to (-1, -1) which we can map to (0,1). getting 0 and 1 in there as well is an issue but it is minor. – fleablood Aug 09 '16 at 05:43
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You can have a bijection from $\{0,1\}^{\mathbb N}$ to $\mathbb P(\mathbb N)$. Take the vector $a\in \{0,1\}^{\mathbb N} $ and map it to $A\subset \mathbb N$ with $i \in A \Leftrightarrow a_i = 1$. (It's easy to show that is a bijection)

Now as $|\mathbb P(\mathbb N)| = |\mathbb R|$, you can construct a bijection from $\{0,1\}^{\mathbb N}$ to $\mathbb R$. Note that I have used $\{0,1\}$ instead of your 2.

Stefan
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