Prove that the Cartesian product of the natural and rational numbers is countably infinite

Asked Aug 27 '20 at 19:20

Active Aug 27 '20 at 19:28

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Prove that $\mathbb{N}\times\mathbb{Q}$ is countably infinite. I understand that I somehow have to show that a bijection exists between $\mathbb{N}$ and $\mathbb{N}\times\mathbb{Q}$, I’m just struggling to see a way of labelling the set in such a way that no elements are missed.

edited Aug 27 '20 at 19:28

Brian M. Scott

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asked Aug 27 '20 at 19:20

user798488

Are you familiar with a bijection between $\mathbb{N}$ and $\mathbb{Q}$? – Keen-ameteur Aug 27 '20 at 19:24
Try to prove that $\mathbb{Q} \times \mathbb{N} \sim \mathbb{N} \times \mathbb{N} = \mathbb{Q}$ – openspace Aug 27 '20 at 19:26
Besides, what has your question to do with combinatorics? – José Carlos Santos Aug 27 '20 at 19:27
@Keen-ameteur by representing the rationals in a grid you can label them to create a surjection, proving that $|\mathbb{Q}|\leq|\mathbb{N}|$. This proves a bijection exists, correct? – user798488 Aug 27 '20 at 19:28
Actually it suffices to show that injections exist both ways, and then invoke Schröder-Bernstein. – celtschk Aug 27 '20 at 19:32

Prove that the Cartesian product of the natural and rational numbers is countably infinite

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