Understanding a step in Baire Category Theorem

Question

I have a (probably stupid) question about the Baire Category Theorem. I am looking at the statement that says that in a complete metric space, the intersection of countable many dense open sets is dense in the metric space. Assume that we have the countable collection of dense open sets $\{U_n\}$ in a complete metric space $X$, and let $x \in X, \epsilon>0$. Since $U_1$ is dense in $X$, there is $y_1\in U_1$ with $d(x,y_1)<\epsilon$. Also, as $U_1$ is open, there is $r_1>0$ with $B(y_1;r_1)\subset U_1$. Then, we can arrange $r_1<1$ such that $\overline{B(y_1;r_1)} \subset U_1\cap B(x;\epsilon)$. Now my question is why we can arrange that the closure will be contained in each of them? I think intuitively it sounds correct, but I didn't succeed to prove it rigorously. Can you please help me here?

Answer 1

$d(x, y_1) < \varepsilon$ actually means $y_1 \in B(x; \varepsilon)$. Also $y_1 \in U_1$ by construction.

Hence $y_1 \in U_1 \cap B(x;\varepsilon)$. Now, since $U_1 \cap B(x;\varepsilon)$ is an open neighbourhood of $y_1$, there exists $r < 1$ such that $B(y_1, r) \subseteq U_1 \cap B(x;\varepsilon)$. Now define $r_1 = \frac{r}2$ and notice

$$\overline{B(y_1, r_1)} = \overline{B\left(y_1, \frac{r}2\right)} \subseteq B(y_1, r) \subseteq U_1 \cap B(x;\varepsilon)$$