Using the fact that $\mathbb N \times \mathbb N$ is countable, or otherwise, prove that the following sets are countable.

a) the set of all points in the $(x,y)$ plane with rational coordinates

b) the set of all $2\times 2$ matrices with integer entries.

I really don't know where to start... and I've never done matrices either!

I know that I need to prove there exists a bijection, but from there I'm lost.

  • 1
    $\begingroup$ For (a), can you use the fact that the rationals are countable? $\endgroup$ – Alexis Olson Nov 2 '16 at 22:07
  • 2
    $\begingroup$ If it helps, you can think of $2\times 2$ matrices as vectors in $\Bbb R^4$ as there is an easy bijection between them. $\begin{bmatrix}a&b\\c&d\end{bmatrix}\leftrightarrow (a,b,c,d)$. The only thing special about matrices is how they add and multiply, but that has nothing to do with the current problem. $\endgroup$ – JMoravitz Nov 2 '16 at 22:09

a) There's two parts to this question. 1. Show there exists a bijection between $\mathbb{N}$ and $\mathbb{Q}$ (a famous result that hopefully you're given). 2. If $f \colon A \to B$ is a bijection and $g \colon C \to D$ is a bijection, then $f \times g \colon A \times C \to B \times D$ defined by $(f \times g)(a, c)=(f(a), g(c))$ is a bijection.

b) For your purposes, you're essentially trying to show $(\mathbb{Z}^2)^2$ is countable. Knowing that $\mathbb{Z}$ is countable, you can apply part 2. to $\mathbb{Z}^2$ to show that $\mathbb{Z}^2$ is countable, and then to show that $(\mathbb{Z}^2)^2$ is countable...well...apply part 2. again.

Fun fact: The insight you get from this problem hopefully shows you most generally that $\prod^n_{i=1} C_i$ for any finite collection $C_1, \dots, C_n$ of countable sets is countable.

| cite | improve this answer | |

For the purpose of just this problem, your $2\times 2$ matrices are just ordered tupls of $4$ integers. That is, they are isomorphic to elements of $\Bbb{N}^4$, which can be written as $\Bbb{N}^2\times \Bbb{N}^2$. Each of those has the required bijection, and now you are down to finding a bijection of $\Bbb{N}\times \Bbb{N}$ to $\Bbb{N}$.

Think about composiing those maps with each other.

For the rationals, it is easy to show that you can have a surjection of the integers to the rationals (choose denominators of 1, for example. The surjection of the rationals to the integers can be made by pretending the numrator and denominator are each independent members of $\Bbb{N}$ so the fraction can be surjectively related to $\Bbb{N}^2$ which in turn is surjective to $\Bbb{N}$.

You need to flesh out these ideas a bit.

| cite | improve this answer | |
  • $\begingroup$ for part a) do i need to show QxQ is countable? I could probably do that, I am just a bit stuck with what it is asking me to do? $\endgroup$ – F.A Nov 2 '16 at 22:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.