union of a Sequence of closed subsets Let $F_1,F_2,...$ be closed subsets of some topological space $X$. If for each $x \in X$ there is a neighborhood of $x$ (open set containing) such that
$N_x \cap F_j \neq \emptyset$
for only finitely many $j$. Prove the arbitrary union of the $F_j$ is closed in $X$.
Here is my attempt and all I have so far,
So I would like to show the complement is open in $X$, that is,
$X \backslash \bigcup_{j=1}^\infty F_j \subset X$
is open. Let $x \in X \backslash \bigcup_{j=1}^\infty F_j$ be arbitrary, we wish to put $x$ in a neighborhood that is disjoint from $\bigcup_{j=1}^\infty F_j$
So we are guaranteed the existence of a neighborhood $N_x$ of $x$ such that
$N_x \cap F_1,...,N_x \cap F_n$
all have non-trivial intersection. But for $F_{n+1},F_{n+2},...$ we know the intersection with $N_x$ is empty, that is
$N_x \cap F_{n+1},N_x \cap F_{n+2}$
are all empty. Also arbitrary union of open is open and since the $F_j$ are closed their complements in $X$ are open, so can I take the complements of the infinitely many $F_j$ that is $X \backslash \bigcup_{k=n+1}^\infty F_k$ is the open neighborhood of $x$? a bit lost. SO basically is it saying that those finitely many $F_j$ have elements that can get close to $x$ but after the threshold of $n$ the $F_k$ become disjoint from balls around $x$? and they are complements of closed so each is open and arbitrary union of open is open?
 A: You started with  a countable union of $F_i$'s. You are supposed to consider uncountable unions also. However, to make my notations consistent with yours I will stick to your countable union. The same proof works for uncountable unions also. Only a minor change in notation is needed.
$N_x \cap F_1^{c}\cap...\cap F_n^{c}$ is an open set containing $x$ and it is contained in $X \setminus \bigcup_n F_n$. This proves that every point $x$ of $X \setminus \bigcup_n F_n$ is an interior point, so $X \setminus \bigcup_n F_n$ is an open set. Hence, $\bigcup_n F_n$ is closed.
A: Let $x \notin \bigcup_{i \in I} F_i$. We know that there is an open neighbourhood $N_x$ of $x$ such that $J:= \{i \in I: F_i \cap N_x \neq \emptyset\}$ is finite.
We also know that $x \notin F_j$ for all $j \in J$, so $O=N_x \bigcap_{j \in J}(X\setminus F_j)$ is an open neighbourhood of $x$, being a finite intersection of such open sets. And $O$ does not intersect any $F_i$: the ones with $i \notin J$ are already missed by $N_x$, so certainly by the smaller $O$, and all the $F_j, j \in J$ are missed by construction. So $O$ misses $\bigcup_{i \in I} F_i$ and as $x$ was arbitrary, this union is indeed closed.
