Is $2^{|\mathbb{N}|} = |\mathbb{R}|$?

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!

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What on earth does "a map of all reals can be done in binary" mean? – Chris Eagle Oct 8 '12 at 18:55
Have you tried searching this site for the answer? – Asaf Karagila Oct 8 '12 at 19:06

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
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.