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What is the cardinality of $\Bbb{N^N}$?

my answer: $|\mathbb{R}|$ $=$$|2^\mathbb{N}|$ $\leqslant$ $|\mathbb{N}^\mathbb{N}|$ $\leqslant$ $|\mathbb{R}^\mathbb{N}|$ $=$ $|(2^\mathbb{N})^\mathbb{N}|$ $=$ $|2^{\mathbb{N}\times\ \mathbb{N}}|$ $=$ $|2^\mathbb{N}|$ $=$ $|\mathbb{R}|$

In the end $|\mathbb{N}^\mathbb{N}|$ $=$ $|\mathbb{R}|$

Is that okay?

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Yes, except that the third item from the end should be $\left|2^{\Bbb N\times\Bbb N}\right|$. – Brian M. Scott Jan 20 '13 at 20:24
Yes, but i didn't know \times\. Thank you – Agenog Jan 20 '13 at 20:27
You’re welcome. – Brian M. Scott Jan 20 '13 at 20:27

Yes. The answer is fine.

(Incidentally my students just have this question in their homework)

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Maybe he is your student ;) – Belgi Jan 20 '13 at 20:52
I doubt that, we gave a slightly different formulation. – Asaf Karagila Jan 20 '13 at 21:04
I was kidding, though I guess you gave your homework in Hebrew so the different formulation can be a result of translation – Belgi Jan 20 '13 at 21:06
Oh, it was more than that. It was different mathematically :-) – Asaf Karagila Jan 20 '13 at 21:13

Your answer is correct, though I would of added that you know that for cardinals $k\leq l,l\leq k\,\implies k=l$

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