Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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|cite|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

1 Answer 1

That is pretty much it. We can write $\mathbb{Q}^k=\{(x_1,...,x_k)\}$ and this is countable because the cartesian product of countable sets is countable ($\mathbb{Q}$ is countable). When constructing your arbitrary neighbourhood, note that we can do the following: Let $y\in\mathbb{R}^k$ and $\epsilon>0$. Set a rational number $x_i$ such that $y-\epsilon<x_i<y+\epsilon,$ with $x=(x_1,...,x_k).$ Then we have \begin{equation*} d(x,y)<\sqrt{\sum^{k}_{i=1}\epsilon^2}=\sqrt{k}\epsilon\to 0 \end{equation*} which completes the proof.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.