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Thread: Nondenumberable

  1. #1
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    Nondenumberable

    I have to prove that $\displaystyle \mathbb{N}^\mathbb{N}$is denumberable; hence, if I understood it right, The set of all mappings from $\displaystyle \mathbb{N}$ to $\displaystyle \mathbb{N}$..

    We worked with these 0,1 tables, which I can work with if I get a mapping, but not if it should work for all mappings.

    Could you help me?
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  2. #2
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    Quote Originally Posted by Frings View Post
    I have to prove that $\displaystyle \mathbb{N}^\mathbb{N}$is denumberable; hence, if I understood it right, The set of all mappings from $\displaystyle \mathbb{N}$ to $\displaystyle \mathbb{N}$..
    We worked with these 0,1 tables, which I can work with if I get a mapping, but not if it should work for all mappings.
    I am not sure that I understand.
    But is this it? You have shown that $\displaystyle 2^{\mathbb{N}}$, the set of all mappings $\displaystyle f:\mathbb{N}\mapsto\{0,1\} $, is uncountable?

    But this question follows from $\displaystyle 2^{\mathbb{N}}\subset \mathbb{N}^{\mathbb{N}}$.
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