# Proof that $\mathbf{R}^n$ has a countable base.

I'm a high school student self-studying analysis, and I'm reading Apostol's book. I wrote a proof that shows that $\mathbf{R}^n$ has a countable base, but am unsure if it is correct. I just wanted some feedback on it. Thanks!

Prove that $\mathbf{R}^n$ has a countable base.

Proof. We construct an open cover $\{V_{\alpha}\}$ for $\mathbf{R}^n$ by considering the union of all neighborhoods with rational centers and rational radii, namely $$\{V_{\alpha}\} = \{ N_q(n):q, n \in \mathbf{Q} \}$$ Pick an arbitrary $x \in \mathbf{R}^n$, and take an arbitrary open set $S$ such that $x \in S.$ Since $S$ is open, there must be a neighborhood $N_{\epsilon}(x)$ with radius $\epsilon$ about $x$ such that $N_{\epsilon}(x) \subset S$.

Take point $j \in \mathbf{Q}^n$ such that $d(j,x) < \frac{\epsilon}{100}$, which exists since $\mathbf{Q}^n$ is dense in $\mathbf{R}^n.$

Now, take $k \in \mathbf{Q}^n$ so that $\frac{\epsilon}{100} < k < \frac{\epsilon}{10}$. Therefore, $N_k(j) \in \{V_{\alpha}\}$. Clearly, $$x \in N_k(j) \subset N_{\epsilon}(x) \subset S.$$ $\{V_{\alpha}\}$ is thus a base.

$\{V_{\alpha}\}$ is countable, since it is a union of a countable collection of countable sets.

Therefore $\{V_{\alpha}\}$ is a countable base.

This is $100\%$ correct, and I'm answering this so the question has an answer.