# Is the closure of $\mathbb Q \times \mathbb Q$ equal to $\mathbb R \times \mathbb R$?

I know the closure of $\mathbb Q$ is $\mathbb R$, but does this imply that the closure of $\mathbb Q \times \mathbb Q$ equal to $\mathbb R \times \mathbb R$?

• To render $\Bbb {R} \times \Bbb {R}$ type $\Bbb {R} \times \Bbb {R}$. To get $\Bbb R^2$ type $\Bbb {R}^2$. For more info on formatting your questions/ answers with MathJax see this tutorial.
– user137731
Commented May 19, 2015 at 3:36
• Commented Jul 17, 2015 at 23:46

In fact , $$\overline{A\times B}=\bar {A}\times \bar{B}$$for any two subsets $A,B\in \mathbb R$.
Here, $$\overline{\mathbb Q\times \mathbb Q}=\bar {\mathbb Q}\times \bar{ \mathbb Q}=\mathbb R\times {\mathbb R}$$
Yes it does. The statement that the closure of $\Bbb Q$ is $\Bbb R$ says that given an element $x$ of $\Bbb R$ you can find a sequence of elements of $\Bbb Q$ that converge to $x$. If you know that, if I give you an element $(x,y)\in \Bbb {R \times R}$ can you find a sequence in $\Bbb {Q \times Q}$ that converges to it?
• Yes. If you want to be formal it takes a bit of doing, just chasing the error bounds. I like to see $\epsilon - \delta$ proofs as challenge-response. You claim your sequence in $\Bbb Q \times Q$ approaches $(x,y)$ I challenge you with an $\epsilon$ and you have to find an $N$ such that all terms after $N$ are within $\epsilon$ of $(x,y)$. You have been told that the sequence you used approaches $x$, so you can challenge whoever gave it to you to find an $N'$ such that all terms after that are within $\epsilon'$ of $x$. If you choose $\epsilon'$ well, and similar for $y$... Commented May 19, 2015 at 4:18