Baire Category Theorem proof confusion

Question

I am not understanding the following proof from Folland.

5.9 The Baire Category Theorem. Let $X$ be a complete metric space.

a. If $\left\{U_{n}\right\}_{1}^{\infty}$ is a sequence of open dense subsets of $X$, then $\bigcap_{1}^{\infty} U_{n}$ is dense in $X$.

b. $X$ is not a countable union of nowhere dense sets.

Proof. For part (a), we must show that if $W$ is a nonemply open set in $X$, then $W$ intersects $\bigcap_{1}^{\infty} U_{n}$. Since $U_{1} \cap W$ is open and nonempty, it contains a ball $B\left(r_{0}, x_{0}\right)$, and we can assume that $0<r_{0}<1$. For $n>0$, we choose $x_{n} \in X$ and $r_{n} \in(0, \infty)$ inductively as follows: Having chosen $x_{j}$ and $r_{j}$ for $j<n$, we observe that $U_{n} \cap B\left(r_{n-1}, x_{n-1}\right)$ is open and nonempty, so we can choose $x_{n}, r_{n}$ so that $0<r_{n}<2^{-n}$ and $\overline{B\left(r_{n}, x_{n}\right)} \subset U_{n} \cap B\left(r_{n-1}, x_{n-1}\right)$. Then if $n, m \geq N$, we see that $x_{n}, x_{m} \in B\left(r_{N}, x_{N}\right)$, and since $r_{n} \rightarrow 0$, the sequence $\left\{x_{n}\right\}$ is Cauchy. As $X$ is complete, $x=\lim x_{n}$ exists. Since $x_{n} \in B\left(r_{N}, x_{N}\right)$ for $n \geq N$ we have $$ x \in \overline{B\left(r_{N}, x_{N}\right)} \subset U_{N} \cap B\left(r_{0}, x_{0}\right) \subset U_{N} \cap W $$$x \in \overline{B\left(r_{N}, x_{N}\right)} \subset U_{N} \cap B\left(r_{0}, x_{0}\right) \subset U_{N} \cap W$ for all $N$, and the proof is complete. As for (b), if $\left\{E_{n}\right\}$ is a sequence of nowhere dense sets in $X$, then $\left\{\left(\bar{E}_{n}\right)^{c}\right\}$ is a sequence of open dense sets. Since $\bigcap\left(\bar{E}_{n}\right)^{c} \neq \varnothing$, we have $\bigcup E_{n} \subset \bigcup \bar{E}_{n} \neq X$.

Why does showing an arbitrary open set intersect $U_N$ is not empty imply $\bigcap_{i=1}^\infty U_n$ is dense in $X$? Can this be phrased in terms of $\overline{\bigcap_{n=1}^\infty U_n}=X$?

Answer 1

$D$ is dense in $X$ iff $\overline{D} = X$. This is the standard definition.

This is equivalent to

for all non-empty open sets $O$: $O \cap D\neq \emptyset$.

For if $O$ is non-empty and $O \cap D = \emptyset$ we have that $D \subseteq X\setminus O$, which is closed so $\overline{D} \subseteq X\setminus O \neq X$ and $D$ would not have $X$ as its closure. For the converse: let $C$ be a closed set that contains $D$. If $C\neq X$ then $X\setminus C$ would be non-empty open and disjoint from $D$, contradiction. So $X$ is the only closed superset of $D$ or $\overline{D}=X$.

This is applied in the proof to $D=\bigcap_n U_n$ and $O=W$. So we need to find a point in $W$ that lies in all $U_n$ and we're done.