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 $\mathbb R \times \mathbb R$?
2 Answers
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In fact , $$\overline{A\times B}=\bar {A}\times \bar{B} $$for any two subsets $A,B\in \mathbb R$.
Here, $$\overline{\mathbb Q\times \mathbb Q}=\bar {\mathbb Q}\times \bar{ \mathbb Q}=\mathbb R\times {\mathbb R}$$
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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?
answered May 19, 2015 at 3:40
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1$\begingroup$ Oh! is it that I can take the sequence that approaches x, and then the sequence approaching y, and stick them together as ordered pairs? And then tht sequence of ordered pairs would approach (x, y) $\endgroup$ Commented May 19, 2015 at 4:09
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$\begingroup$ 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 $y$... $\endgroup$ Commented May 19, 2015 at 4:18
$\Bbb {R} \times \Bbb {R}$. To get $\Bbb R^2$ type$\Bbb {R}^2$. For more info on formatting your questions/ answers with MathJax see this tutorial. $\endgroup$