Let $X$ be a metric space. If $F_i \subset X$ is closed for $1 \leq i \leq n$, prove that $\bigcup_{i=1}^n F_i$ is also closed.
I'm looking for a direct proof of this theorem. (I already know a proof which first shows that a finite intersection of open sets is also open, and then applies De Morgan's law and the theorem "the complement of an open set is closed.") Note that the theorem is not necessarily true for an infinite collection of closed $\{F_\alpha\}$.
Here are the definitions I'm using:
Let $X$ be a metric space with distance function $d(p, q)$. For any $p \in X$, the neighborhood $N_r(p)$ is the set $\{x \in X \,|\, d(p, x) < r\}$. Any $p \in X$ is a limit point of $E$ if $\forall r > 0$, $N_r(p) \cap E \neq \{p\}$ and $\neq \emptyset$. Any subset $E$ of $X$ is closed if it contains all of its limit points.
Best Answer
Let $F$ and $G$ be two closed sets and let $x$ be a limit point of $F\cup G$. Now, if $x$ is a limit point of $F$ or $G$ it is clearly contained in $F\cup G$. So suppose that $x$ is not a limit point of $F$ and $G$ both. So there are radii $\alpha$ and $\beta$ such that $N_\alpha(x)$ and $N_\beta(x)$ don't intersect with $F$ and $G$ respectively except possibly for $x$. But then if $r=min (\alpha,\beta)$ then $N_r(x)$ doesn't intersect with $F\cup G$ except possibly for $x$, which contradicts $x$ being a limit point. This contradiction establishes the result. The proof can be extended easily to finitely many closed sets. Trying to extend it to infinitely many is not possible as then the "min" will be replaced by "inf" which is not necessarily positive.