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I'm having trouble proving that the sets $\{0,1\}^\mathbb{N}$ and $\mathbb{N}^\mathbb{N}$ have different cardinalities.

What I tried to do: I assumed there exists a function $F:$ $\{0,1\}^\mathbb{N}\to$ $\mathbb{N}^\mathbb{N}$ which is onto $\mathbb{N}^\mathbb{N}$. I then tried using Cantor's diagonal argument to disprove the claim, however, I cannot find (or build) a function which belongs to $\mathbb{N}^\mathbb{N}$ but doesn't belong to the image of $F$. To elaborate: Since $\{0,1\}^\mathbb{N}$ is uncountable, I assumed there is an index set $I$ which is equivalent to it. Then, for each $i\in I$ there is a function $f_i\in\{0,1\}^\mathbb{N}\to$ $F(f_i)=m_i$ where $m_i\in \mathbb{N}^\mathbb{N}$ (Assuming that $I$ is equivalent to $\mathbb{N}^\mathbb{N}$) . How can I build a function $b\in \mathbb{N}^\mathbb{N}$ that is not equal to all $m_i,\ \ i\in I$ using Cantor's diagonal argument? Is there a specific choice for $I$ which would make the problem easier?

Would appreciate any help and hints.

Rory Daulton
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Dylan132
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    What does equivalent mean? – Andrés E. Caicedo Nov 29 '15 at 14:45
  • I haven't learned about cardinality yet, but I'm trying to prove that the cardinalty of {0,1}^N is smaller than the cardinality of N^N, or that there doesn't exist a one-to-one function onto N^N (whose domain is {0,1}^N) – Dylan132 Nov 29 '15 at 14:50
  • Why are you trying to prove this? (Meaning: Why do you think this is true?) – Andrés E. Caicedo Nov 29 '15 at 14:51
  • The question was to check whether both sets are not equivalent. I tried to prove it for I had no idea regarding how to disprove it. If they are equivalent, how can I prove that? (if you could give me a hint) – Dylan132 Nov 29 '15 at 15:00
  • See http://math.stackexchange.com/questions/1093259/power-of-sets-0-1-mathbbn-simeq-mathbbn-mathbbn and other related questions. You can find some of them here, here or here. – Martin Sleziak Nov 30 '15 at 09:29

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Note that $\Bbb{N^N}$ is a subset of $\mathcal P(\Bbb{N\times N})$. Now use the fact that $\mathcal P(\Bbb{N\times N})$ and $\mathcal P(\Bbb N)$ have the same cardinality.

Asaf Karagila
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