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Use the unique prime factorization property of $\mathbb{Z}$ (fundamental theorem of arithmetic) and the Schroeder-Bernstein theorem to show that

$$| \mathbb{N} | = | \mathbb{N} \times \mathbb{N} |.$$

Anyone knows how to start this question?

I know that Schröder–Bernstein theorem, states that, if there exist injective functions f : A → B and g : B → A between the sets A and B, then there exists a bijective function h : A → B.

But then I don't know how to apply it to this question and write a detailed approach to it. Help would be appreciated.

quid
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user271716
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1 Answers1

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One can proceed like this:

  • The map $n \mapsto (n,0)$ is injective from $\mathbb{N}$ to $\mathbb{N}^2$.

  • The map $(m, n) \mapsto 2^m 3^n$ is injective from $\mathbb{N}^2$ to $\mathbb{N}$ (here you need to use uniqueness of prime factorization).

  • Thus by Schröder–Bernstein there is a bijection.

quid
  • 42,135
  • In case your natural number do not include $0$ just replace $0$ by $1$. – quid Oct 27 '15 at 16:44
  • i know how to prove from N to N^2 is injective, is there a detailed explanation for the map (m,n)↦2^m3^n is injective from ℕ^2 to N? – user271716 Oct 27 '15 at 17:10
  • Sorry for the delay in responding. The explanation that this map is injective is basically the funcdamental thm of arith. Suppose the image of $(m,n)$ and $(m';n')$ are the same, call it $a$. Then $a= 2^m3^n$ and $a= 2^{m'}3^{n'}$. However by FTA $a$ has a unique factorization into primes so $m=m'$ and $n = n'$. – quid Oct 28 '15 at 13:13