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.
Best Answer
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 $y_i - \epsilon < x_i < y_i + \epsilon$, 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.