If $A$ is a countable set, the cartesian product $A \times A$ is countable.

Question

Prove that if $A$ is a countable set, then the Cartesian product $A \times A$ is also countable.

First suppose that A is countable.

If A is finite, then since $ A \neq \emptyset $ there exists an integer $n$ and a bijection $f : A → \Bbb N$.

In particular, f is a one-to-one mapping of $A$ into $ \Bbb N$. So we have found our $f$, if A is finite.

If A is infinite, then $A$ is countably infinite. Therefore, there is a bijection $f : A → \Bbb N$. Thus, in both cases, we have a one-to-one mapping $f : A → \Bbb N$. Now suppose that we have a one-to-one mapping $f$ of $A$ into $ \Bbb N$. Then $f$ maps $A$ onto its range. Therefore $A ≈ran( f)$. But $ran(f)$ is a subset of $A$. So $A$ is countable, as every subset of a countable set is countable.

We show that $ A \times A $ is countable by defining a function $ g: A \times A → A $ explicitly.

So, let $ g(a,b) = 2^{g(a)} 3^{g(b)} $ for all $ (a, b) ∈ A × A $.

This is clearly a well- defined function and the function is not onto, since a number like 7 is not in the range. So we will not try to show that f is a bijection between $ A \times A $ and $A$.

Since the prime factorization of a natural number is unique, the function is one-to-one. Thus we have a one-to-one mapping $ f : A × A → A $.

We have shown that $ A × A$ is equivalent to $A$, thus $ A \times A$ is also countable.

You seem to be assuming $A\subset\mathbb{N}$. What I would do is define an injective $g:A\to\mathbb{N}$ (exists since $A$ is countable) and define $f(a,b)=2^{g(a)}3^{g(b)}$. Since $g$ is injective, the rest of the proof works — Math101, Oct 09 '22 at 18:19
So define an injective $g : A → \Bbb N$ instead of the bijection: $ f : A → 1,2,...,n$. Could I then remove the part ''If $A$ is finite.......every subset of a countable set is countable'' completely? — Allison, Oct 09 '22 at 18:27
I tried editing the post with your feedback, but Im not really sure if I'm doing it correctly . Could you edit my post so I mark it as answered? — Allison, Oct 09 '22 at 18:40
Does this answer your question? The cartesian product $\mathbb{N} \times \mathbb{N}$ is countable — Qiaochu Yuan, Oct 09 '22 at 19:31
Thanks Yuan @QiaochuYuan, I read it , but now in my proof I believe I'm assuming $ A ⊂ \Bbb N$ and I dont know how to fix this. Did I do it right? — Allison, Oct 09 '22 at 19:33

Kurt G. · Accepted Answer · 2022-10-11T19:20:02.733

Too long for a comment.

You do not have to distinguish the cases $A$ finite and $A$ (countably) infinite.
When you assume $A$ is finite I do not see how you get a bijection $f:A\to\mathbb N$. That's impossible when $A$ is finite.
Even if you get an injection $f:A\to\mathbb N$ you have not found "our" $f$ that helps proving that $A\times A$ is countable.
The above list of errors could be extended but I stop here.
GBA has given the hint that allows to prove the countability of $A\times A$ in just a few lines:
Since $A$ is countable we can consider it as a subset of $\mathbb N$. Then take two primes, say, $2$ and $3$ and get an injection from $A\times A$ to $\mathbb N$ by $$ g(a,b)=2^a3^b\,. $$ By uniqueness of prime factorization there cannot be distinct pairs $(a,b)$ and $(a',b')$ such that $g(a,b)=g(a',b')$. So this is an injection.
In other words, we have attached to every $(a,b)\in A\times A$ a natural number $n\in\mathbb N$ that characterizes $(a,b)$ uniquely. Thus, $A\times A$ is countable.

If $A$ is a countable set, the cartesian product $A \times A$ is countable.

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