Showing that a locally finite collection preserves the closure

Question

I managed to show (a), but am stuck at (b). I think I have to manipulate the definition of local finiteness but cannot find a way. Could anyone please help me with (b)?

Answer 1

First note that $$ \bigcup_{X\in\mathcal{X}} \overline{X} \subseteq \overline{\bigcup_{X\in\mathcal{X}} X}, $$ since each $X$ is contained in the union, so it's closure is contained in the closure of the union. Thus we just need to prove the other inclusion.

For the other direction, it suffices to prove that the union of the closures is closed, since the union of the $\overline{X}$s contains the union of the $X$s, so if it's closed itself, it contains the closure of the union of the $X$s.

Let $$x\not\in \bigcup_{X\in\mathcal{X}}\overline{X}.$$ By local finiteness of $\{\overline{X}\}$, let $U$ be a neighborhood of $x$ intersecting only finitely many sets $\overline{X_i}$ for $1\le i \le n$. Then $$U\cap \bigcap_{i=1}^n \overline{X_i}^C$$ is an open neighborhood of $x$ disjoint from $\bigcup_{X\in\mathcal{X}}\overline{X}.$ Thus the complement of $\bigcup_{X\in\mathcal{X}}\overline{X}$ is open, or $\bigcup_{X\in\mathcal{X}}\overline{X}$ is closed, as desired. Thus we have the reverse inclusion. Hence $$ \bigcup_{X\in\mathcal{X}} \overline{X} = \overline{\bigcup_{X\in\mathcal{X}} X}, $$ as desired.