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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!

Proof:

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.

Q.E.D.

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    $\begingroup$ Correctamundo.${}$ $\endgroup$ – Pedro Tamaroff Jan 31 '14 at 5:53
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    $\begingroup$ 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. $\endgroup$ – Moishe Kohan Jan 31 '14 at 6:29
  • $\begingroup$ Why didn't you accept the answer given? $\endgroup$ – Viktor Glombik Apr 30 '19 at 19:16
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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.

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