# A problem on separabale metric space

I want to prove that $R^k$ has a countable dense (any point in $R^k$ is either a limit point of that subset or contains in that subset) subset. Shall I use induction on $k$ ?

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Think: what is a countable dense subset of $\mathbb{R}$? How can you easily generalize this to $\mathbb{R}^n$? – Cameron Williams Sep 6 '13 at 15:38
@CameronWilliams: For $\Bbb R$ it is $\Bbb Q$. So, for $\Bbb R^k$ it will be set with rational co-ordinates. But how do I prove it ? – aaaaaa Sep 6 '13 at 15:42
If $(x_1,x_2,...,x_k)$ is a real vector, then you can arbitrarily approximate each coordinate with rational numbers, so we can choose a rational vector $(r_1,...,r_k)$ such that $|x_i-r_i|\lt\frac{\varepsilon}k$. Now it suffices to note that $||\mathscr{x}-\mathscr{r}||_2\le k||\mathscr{x}-\mathscr{r}||_{\infty}$. – walcher Sep 6 '13 at 15:55

If you want to use induction, you will have to show that a) $\mathbb{R}$ is separable and b) that the product of two separable spaces is again separable.