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I'd like to prove that $\mathrm{card}(2^{\mathbb{N}})=\mathrm{card}(\mathbb{N}^\mathbb{N})$, I have the following 'sketch' but I'm not sure if this works.

$|2^{\mathbb{N}}|\leq|\mathbb{N}^{\mathbb{N}}|\leq|2^\mathbb{N^{\mathbb{N}}}|=|2^{\mathbb{N}\times\mathbb{N}}|=|2^\mathbb{N}|$, then $|2^{\mathbb{N}}|=|\mathbb{N}^\mathbb{N}|$

I'm taking for granted the first inequality, (i.e: $|2^{\mathbb N}|\leq|\mathbb{N}^{\mathbb{N}}|$), could be done a further proof about this. Would it be enough to point out that the functions in $2^{\mathbb{N}}$ are in $\mathbb{N}^{\mathbb{N}}$ but there are functions in the last one that are not in the first one? Should I try to give a more formal proof?

Martin Sleziak
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Cure
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    You can just use that if $|A| \le |B|$, then $|A^C| \le |B^C|$. – Stephen Montgomery-Smith Apr 12 '14 at 03:59
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    You must use parenthesis when writing two exponents. Otherwise it's ambiguous and possibly wrong. – Asaf Karagila Apr 12 '14 at 04:02
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    One and two and three, and there are most definitely a few more. I even recall giving an outline of an explicit bijection once. – Asaf Karagila Apr 12 '14 at 04:05
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    @Asaf You posted (probably by mistake) one of the links twice. I am adding another one: http://math.stackexchange.com/questions/183513/what-is-aleph-0-powered-to-aleph-0 (Since there are plenty of questions like this, some of them could be closed as duplicates.) – Martin Sleziak Apr 12 '14 at 07:26
  • @Martin: Yes, thank you. As for duplicate, I think we should probably pay attention to the content of the question itself. Even if the question is the essentially the same, it might be slightly different -- enough not to merit closure. – Asaf Karagila Apr 12 '14 at 07:28

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I'll just point out that when you write $2^{\mathbb N^{\mathbb{N}}}$ , there is a certain ambiguity to it.

To make sure that it is clear what you have in mind, you should write $(2^{\mathbb N})^{\mathbb{N}}$. (This is what you have used here.)

The other possible meaning is $2^{(\mathbb N^{\mathbb{N}})}$.

In fact, when someone writes $a^{b^c}$, they usually mean $a^{(b^c)}$. (If someone wants to write $(a^b)^c$, they can write $a^{bc}$ instead. See How to evaluate powers of powers (i.e. $2^3^4$) in absence of parentheses? or $x^{y^z}$: is it $x^{(y^z)}$ or $(x^y)^z$?

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Martin Sleziak
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Your proof of inclusion is fine. You have shown that each element of $2^{\Bbb N}$ is also an element of $\Bbb N^{\Bbb N}$.

Ross Millikan
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