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What is $\left|\mathbb{N}^\mathbb{N}\right|$?

I know that $\forall n\in\mathbb{N}: \left|\mathbb{N}^n\right|=\left|\mathbb{N}\right|$. But is this also true for the limit?

  • No, it is not countable because it contains $2^{\mathbb{N}}$ (and it is not a limit in the sense you probably mean it). – Tobias Kildetoft May 24 '16 at 11:18
  • But then, assuming the continuum hypothesis, would it be $\aleph_1$? Or even bigger? – C. Maier May 24 '16 at 11:23
  • Cardinality doesn't play nice with "limits". Indeed as Tobias has said, $\mathbb{N}^\mathbb{N}$ contains $2^\mathbb{N}$ which is not countable by Cantor's theorem. There's a nice example of cardinality not playing nice with limits that you might want to think about. Consider three bins, A, B, C. The first bin has infinitely many balls in it, indexed by the natural numbers. The other two are empty. Now in order, we move two balls from A to B, and then one ball (still in order) from B to C. "Continue until A is empty" (pretend that makes sense for a moment). How many balls are in B now? – Ian May 24 '16 at 11:23
  • @C.Maier Don't worry about CH at your level. It turns out that it is the same as $\mathfrak{c}$, also known as $|\mathbb{R}|$, also known as $\beth_1$. Whether this is $\aleph_1$ is irrelevant to your current purposes. – Ian May 24 '16 at 11:24
  • Woops. Meant to link to this one: http://math.stackexchange.com/q/110211/10014 – Najib Idrissi May 24 '16 at 11:24

1 Answers1

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It it true for every $n$ cause you can find a bijection between both sets. However, it is not true for the limit case. The canonical way to see this is to find a injection from $[0,1]$ to $\mathbb{N}^\mathbb{N}$. There is a well known argument writing $x$ in $[0,1]$ in binary base that allows you to inject $[0,1]$ in $2^\mathbb{N}$ and then it's easy to inject $2^\mathbb{N}$ in $\mathbb{N}^\mathbb{N}$. This gives you that the cardinal of $\mathbb{N}^\mathbb{N}$ is at least the cardinal of $[0,1]$; the cardinal of the real numbers (use arctan function to see it), usually denoted by $c$. As $c>|\mathbb{N}|$ you are done.

  • How do you construct the bijection that you are talking about? To each $x \in [0,1]$ you associate what from $\Bbb N ^{\Bbb N}$? – Alex M. May 24 '16 at 17:22
  • @AlexM. Better alex? U want me to explain? I will have to begin to pay you for your hourly dose of wisdom and bliss. I know! U want me to upvote your answers so that you have more points? ;) –  May 24 '16 at 19:13
  • No, the problem is that I believe that your argument is mistaken, and I'm not joking or harassing you, please stop believing this. So, could you please edit your answer in order to include an explicit description (or at least a sketch) of this bijection? Also, please stop thinking about me, think about the OP and any future readers who might stumble upon this page (say, redirected from a search engine). – Alex M. May 24 '16 at 19:15
  • @AlexM. You should inform youself beefore speaking. Havent u see that I changed it? Is still mistaken my argument? It was not that bad givem that I wrote the answer in a moment and froom my phone and given that the moment y posted it the question was declared duplicated dont u think? And please... We are almost acquaintances. Stop boring me with your empty peroration... Although, being honest, I appreciate yoou seek for accurateness. Almost makes me forget how sorry I feel for u. Kisses –  May 24 '16 at 19:26
  • Indeed, I hadn't noticed your edit. The present version is correct. – Alex M. May 24 '16 at 19:40