# Cardinality of the set of all natural sequences is $2^{\aleph_0}$ [duplicate]

I was wondering how can you prove that $\mathbb{N}^\mathbb{N} \sim 2^\mathbb{N}$ (where $\mathbb{N}^\mathbb{N}$ is the set of all functon $f:\mathbb{N}\rightarrow \mathbb{N}$).

I think I can show that $2^\mathbb{N} \sim n^\mathbb{N}$ for some $n\in\mathbb{N}$ with a tedious bijection, but I'm not sure what to do with $\mathbb{N}^\mathbb{N}$.

I apologize if the question lacks details/notation, I have only studied elementary set theory and we haven't mentioned $\mathbb{N}^\mathbb{N}$'s cardinality at all (I usually ended up finding injections/bijections to/from $2^\mathbb{N}$, but now I'm curious as to why the above correct).

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## marked as duplicate by Asaf Karagila, Ross Millikan, Sujaan Kunalan, some1.new4u, mrfDec 3 '13 at 21:35

There are probably a dozen other duplicates. –  Asaf Karagila Dec 3 '13 at 20:41
I have searched before for duplicates. If I had stumbled upon the link you posted, then perhaps I didn't recognize the resemblance between the two questions. This could probably mean that the set theory I know is too elementary for this question. Are you saying that the equivalence of cardinalities in question can't be shown using a relatively simple mapping, but require games with "arithmetic of cardinal numbers"? –  Py42 Dec 3 '13 at 20:53
They can. Although the formula is not very simple. –  Asaf Karagila Dec 3 '13 at 21:01
Could you please point me to the said formula? Also, are the tools shown in the possible duplicates are something that is taught in introductory set theory (undergraduate) course? –  Py42 Dec 3 '13 at 21:06
My answer is using only intro course material (the part about $\aleph_0$ anyway). At least if it covers cardinal arithmetics. But as I said, there are probably a dozen more duplicates and at least some of them might be better fitting for you. That's just the first one I found. In either case I have added a description of a bijection between the two sets. –  Asaf Karagila Dec 3 '13 at 21:19

For every number $n\in\Bbb N$ let $\sigma_n$ be the binary sequence of $1$'s of length $n$ and then a $0$, this is - if you want to think of it that way - a representation of $n$ in base $1$. With a zero afterwards.

Now given a sequence $a=\langle a_n\mid n\in\Bbb N\rangle$ where $a_n$ is a natural number, we define the following binary sequence by induction.

1. $b_0$ is the empty sequence.
2. If $b_n$ was defined, $b_{n+1}$ is the concatenation of $b_n$ with $\sigma_{a_n}$. That is we write $b_n$, and then $\sigma_{a_n}$.

$b$ is the sequence we have at the end of the induction process. Then $b\in 2^{\Bbb N}$, which is easy to see, as it is an infinite sequence of $0$-$1$ digits.

And I claim that the map sending $a$ to $b$ (where $a$ is a sequence of natural numbers, and $b$ is a sequence defined as above from $a$) is a bijection.

For this we just observe that the function decoding sequences of $1$'s between $0$'s is an inverse function. It's a bit harder to write down formally, but I'll try.

Let $b=\langle b_n\mid n\in\Bbb N\rangle$ be an infinite binary sequence. We define a sequence of integers.

1. $a_0=\min\{k\mid b_n=0\}$.
2. If $a_0,\ldots,a_n$ were defined, let $t$ be the least $k$ such that $b_k=0$ and $k\geq\sum (a_i+1)$ then $a_n=t-\sum (a_i+1)$.

I might be off with the indices in that second part, but that's the general idea. We find out how to decompose $b$ into pieces looking like $\sigma_k$, then we take the $k$ from each piece.

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