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I'd like to show that $\mathbb{R}^k$ is separable. (A metric space is called separable if it contains a countable dense subset.)

Here's what I have and I'd like to confirm with everyone to see if there's anything else I need to add. Thanks in advance!


The metric space $\mathbb{R}^k$ clearly contains $\mathbb{Q}^k$ as a subset. We know that $\mathbb{Q}^k$ is countable from theorem 2.13 (Rudin, Principles of Mathematical Analysis 3rd Edition, Pg. 29).

To prove that $\mathbb{Q}^k$ is dense in $\mathbb{R}^k$, we need to show that every point in $\mathbb{R}^k$ is a limit point of $\mathbb{Q}^k$.

Let $a=(a_1,a_2,...,a_k)$ be an arbitrary point in $\mathbb{R}^k$ and let $N_r(a)$ be an arbitrary neighborhood of $a$. Let $b=(b_1,b_2,...,b_k)$ where $b_i$ is chosen to be a rational number such that $a_i<b_i<a_i+\frac{r}{\sqrt{k}}$ (this is possible because of theorem 1.20(b)) (Rudin, Pg. 9). The point $b$ is clearly in $\mathbb{Q}^k$, and

$d(a,b) = \sqrt{(a_i-b_i)^2 + ... + (a_k-b_k)^2} < \sqrt{\frac{r^2}{k} + ... + \frac{r^2}{k}} = \sqrt{\frac{kr^2}{k}} = r$

This shows that every point in $\mathbb{R}^k$ is a limit point of $\mathbb{Q}^k$, which completes the proof that $\mathbb{R}^k$ is separable.


share|improve this question
Correctamundo.${}$ –  Pedro Tamaroff Jan 31 '14 at 5:53
I think, you are working too hard: If $Y_i\subset X_i$, $i=1,...,n$ are dense subsets for each $i$, then product of $Y_i$'s is dense in the products of $X_i$'s. This is true for arbitrary topological spaces. –  studiosus Jan 31 '14 at 6:29
Your proof looks fine for me! –  user124187 Feb 1 '14 at 20:49

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