Understanding the proof of intersection of closed sets is closed

Question

Prove for any collection of $\{F_\alpha\}$ of closed sets, $\bigcap_{\alpha} F_{\alpha}$ is closed.

I am reading Methods of real analysis by Goldberg and I am trying to understand the proof of the above stated theorem.

I understand the proof using De Morgan's law. For e.g., this one: using demorgan's law

Goldberg book's proof:

I am having trouble understanding the second sentence:

Let $x \in \left(\bigcap_{\alpha} F_{\alpha}\right)^c$, then any open ball $B_r(x)$ contains a point $y \in \bigcap_{\alpha} F_{\alpha}$...

How can we conclude any open ball contains a point $y$?

Answer 1

If there is an $r$ such that $B[x;r]\cap \bigcap_{F\in \cal F}F=\emptyset$ this would mean that $x$ is in the inner of the complement of $\bigcap_{F\in \cal F}F$, since $B[x;r]\subseteq (\bigcap_{F\in \cal F}F)^{\mathrm complement}$. Thus cannot belong to the closure of $\bigcap_{F\in \cal F}F$.

Answer 2

Suppose $x \in \overline{\bigcap \{F : F \in \mathcal{F}\}}$, so the closure of the intersection. Fix $F_0 \in \mathcal{F}$ arbitarily. Let $B(x,r)$ be any ball around $x$. From being in this closure we know that $B(x,r) \cap \bigcap \{F : F \in \mathcal{F}\} \neq \emptyset$, say they intersect in $y$, but as $y$ is in the intersection, in particular $y \in B(x,r) \cap F_0$. So $x \in \overline{F_0}=F_0$ ($F_0$ is closed). As $F_0 \in \mathcal{F}$ was arbitary, $y \in \bigcap \{F : F \in \mathcal{F}\}$. So $\bigcap \{F : F \in \mathcal{F}\}$ is closed, as it contains every point of it closure.

Essential background knowledge: $\overline{A}=\{x\mid \forall r>0: B(x,r) \cap A \neq \emptyset\}$ and $A$ is closed iff $\overline{A} \subseteq A$.