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

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

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