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I know that $\mathbb{Q}$ is dense in $\mathbb{R}$. What is the next step to prove that $\mathbb{Q}^2$ is also dense in $\mathbb{R}^2$?
Any hints are welcomed.

Sarah. N
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    Do you know that every non-empty open set in $\Bbb R^2$ contains an open set of the form $(a,b)\times(c,d)$, where $a,b,c,d\in\Bbb R$, $a<b$, and $c<d$? – Brian M. Scott Oct 11 '14 at 06:38
  • An alternate hint: A sequence $(a_n, b_n)$ converges in $\mathbb{R}^2$ to $(a, b)$ iff $a_n \to a$ and $b_n \to b$. – Travis Willse Oct 11 '14 at 06:39

2 Answers2

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Hints: (Just as Brian comments)

Every non-empty open set in $\Bbb R^2$ contains an open set of the form $(a,b)\times (c,d)$ and $(a,b)$ contains one element of $\Bbb Q$ at least, say $x$; for the same reason, $(c,d)$ contains $y\in \Bbb Q$. And hence It is easy to see that $\Bbb Q^2$ is dense in $\Bbb R^2$.

drhab
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Paul
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Let $x \in \mathbb{R}^2$, and let $q_n, r_n \in \mathbb{Q}$ such that $q_n \to x_1, r_n \to x_2$.

Then $\|x-(q_n,r_n)\| = \sqrt{(x_1-q_n)^2+(x_2-r_n)^2} \to 0$, hence the desired result.

copper.hat
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