# 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 $$\mathbb{R}^n$$ has a countable base, but am unsure if it is correct. I just wanted some feedback on it. Thanks!

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

Proof. We construct an open cover $$\{V_{\alpha}\}$$ for $$\mathbb{R}^n$$ by considering the union of all neighborhoods with rational centers and rational radii, namely $$\{V_{\alpha}\} = \{ N_q(n):q, n \in \mathbb{Q} \}$$ Pick an arbitrary $$x \in \mathbb{R}^n$$, 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 $$j \in \mathbb{Q}^n$$ such that $$d(j,x) < \frac{\epsilon}{100}$$, which exists since $$\mathbb{Q}^n$$ is dense in $$\mathbb{R}^n.$$

Now, take $$k \in \mathbb{Q}$$ 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.

Thus, $$\{V_{\alpha}\}$$ is a countable base.

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