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I am trying to show whether or not $\mathbb N ^5$ is a countable set. So i need to find a bijective function such that $f:\mathbb N^5 \to \mathbb N$, but I have no clue how to do this, any hints?

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    $\begingroup$ If by $\mathbb{N}^5$ you mean the cartesian power of $\mathbb{N}$, then yes, since the cartesian product of two countable sets is countable (and by induction 5 sets). $\endgroup$
    – megas
    Mar 5, 2015 at 9:13
  • $\begingroup$ I would recommend trying to find two injections and using the cantor bernstein theorem. $\endgroup$
    – DRF
    Mar 5, 2015 at 9:14

2 Answers 2

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Let $g \colon \mathbb N^2 \to \mathbb N$ be any bijection, for example the Cantor pairing $$ g(n,m) = \frac 12 (n+m)(n+m+1) + m $$ Now define $$ f(n_1, n_2, n_3, n_4, n_5) = g\biggl(g\Bigl(g\bigl(g(n_1, n_2), n_3\bigr), n_4\Bigr), n_5\biggr) $$

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By the fundamental theorem of arithmetic, the function $f:\mathbb{N}^5\to\mathbb{N}$ defined by

$$f(n_1,n_2,n_3,n_4,n_5)=2^{n_1}3^{n_2}5^{n_3}7^{n_4}11^{n_5} $$ is injective. Also the function $g:\mathbb{N}\to \mathbb{N}^5$ defined by $n\mapsto (n,0,0,0,0)$ is injective.

With Schroder-Bernstein theorem the two sets $\mathbb{N}$ and $\mathbb{N}^5$ are equinumerous. Hence you get countably infinite.

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