If $A$ and $B$ are countable sets, show that $A \times B$ is countable

Question

My question is If $A$ and $B$ are countable sets, show that $A \times B$ is countable.

I know the definitions to be a countable set are:

A set $A$ is countable if $A$ is finite or countably infinite.
A set $A$ is countably infinite if $A \thicksim \mathbb{N}$ ( same cardinality)

I just want to be walked through this problem for studying reasons.

In resumen, you must show that exists a bijection $f:\Bbb N\times\Bbb N\to\Bbb N$. — Masacroso, Sep 09 '16 at 23:10
... and bijection between $\mathbb{N}$ and $\mathbb{N}\times\mathbb{N}$ and many others. — , Sep 09 '16 at 23:19
@WillO He does not need it: he has just two functions to choose. You need AC to prove that the disjoint union of countably many countable sets is countable. — , Sep 09 '16 at 23:21
@WillO This does not require the Axiom of Choice (or any weaker version stronger than Axiom of Finite Choice). — Hayden, Sep 09 '16 at 23:22
See Cantor proof on the carnality of the rational numbers. Same arguments can be applied. — Doug M, Sep 09 '16 at 23:34

score 0 · Answer 1 · answered Sep 09 '16 at 23:47

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If $A$ and $B$ are countable, then they can be listed $A={a_1, a_2,\ldots}$ and $B={b_1,b_2,\ldots}$. Then the map $(a_k,b_n) \rightarrow 2^k3^n$ is an injection from $A\times B$ into $\mathbb{N}$.

answered Sep 09 '16 at 23:47

B. Goddard

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score 0 · Answer 2 · answered Sep 10 '16 at 00:13

If $A= \{a_1,a_2,... \}$ and $B= \{b_1,b_2,... \}$

Let $A_k=\{(a_n,b_k):n \in \Bbb N \}$ and $k=1,2,...$

I think it's clear that $A_k$ are equinumerous to $A$ for all $k=1,2,...$ and therefore countable

This way $A \times B= \bigcup_{k=1}^{\infty}A_k$ and we know that a countable union of countable sets is a countable set

If $A$ and $B$ are countable sets, show that $A \times B$ is countable

2 Answers2