Closure of union of locally finite collection of subsets equal to union of closures.

Question

Say $\mathcal{S}=\{S_i|i\in I\}$ is a locally finite collection of subsets of a topological space $X$. Then $$\overline{\bigcup_{i\in I} S_i} = \bigcup_{i\in I}\overline{S_i}.$$

I don't see why this is true.. Hope someone can explain it to me!

Answer 1

This is a partial solution that works in spaces where closure coincides with sequential closure, such as Frechet spaces.

Pick $$x\in \overline{\bigcup_{i\in I} S_i}$$

Then because $\mathcal{S}$ is locally finite there is a neighbourhood $x\in U$ such that $U$ has a non-empty intersection with only finitely many $S_i$'s.

Now pick a sequence $x_k\in \bigcup_{i\in I} S_i$ that converges to $x$. You can choose the $x_k$ to lie in $U$ because it is a neighbourhood of $x$ (or in any case they will all lie in $U$ except finitely many of them).

Since there are infinitely many $x_k$ but finitely many $S_i$ where they can possibly lie, by the pigeonhole principle, there is a subsequence that lies entirely in one and the same $S_i$. It follows that $x$ lies in the closure of that particular $S_i$.

Therefore we have proved $$\overline{\bigcup_{i\in I} S_i}\subset \bigcup_{i\in I} \overline{S_i}$$

The other direction is trivial.

Answer 2

Suppose $x$ belongs to the closure of $S=\bigcup_{i\in I}S_i$. Fix a neighborhood $U_0$ of $X$, so $U_0\cap S\ne\emptyset$. On the other hand, $U_0$ can be chosen to only intersect a finite number of the $S_i$, say $S_{i_1},\dots,S_{i_n}$. Any neighborhood $U$ of $x$, $U\subseteq U_0$, must intersect $S$, so $$ U\cap (S_{i_1}\cup\dots\cup S_{i_n})\ne\emptyset $$ because $$ U\cap\biggl(\,\bigcup_{i\in I\setminus\{i_1,\dots,i_n\}}S_i\biggr)=\emptyset $$ In particular, since the neighborhoods of $x$ contained in $U_0$ form a base, $$ x\in\overline{S_{i_1}\cup\dots\cup S_{i_n}}= \overline{S_{i_1}}\cup\dots\cup \overline{S_{i_n}}\subseteq \bigcup_{i\in I}\overline{S_i} $$ The converse inclusion is obvious.

Answer 3

For the "trivial" inclusion: note that for every $i\in I$ you have $S_i\subseteq\bigcup_{i\in I}S_i$ so $\overline{S_i}\subseteq\overline{\bigcup_{i\in I}S_i}$. Taking the union on the left side, we have $\bigcup_{i\in I}\overline{S_i}\subseteq\overline{\bigcup_{i\in I}S_i}$.

For the converse, it is enough to prove that $\bigcup_{i\in I}\overline{S_i}$ is closed. (This is because it trivially contains $\bigcup_{i\in I}S_i$, so it must then contain its closure (i.e. $\overline{\bigcup_{i\in I}S_i}$).)

For that purpose, pick $x\not\in \bigcup_{i\in I}\overline{S_i}$. Due to local finiteness, there is an neigbourhood / open set $U, x\in U$ such that $U$ intersects only finitely many $S_i$'s: let's say $S_{i_1}, S_{i_2},\ldots,S_{i_n}$. Now create a new neighbourhood $U'=U\setminus(\overline{S_{i_1}}\cup\overline{S_{i_2}}\cup\cdots\cup\overline{S_{i_n}})$. It is an open set, it still contains $x$, and it does not intersect any of $S_i$'s, which implies (as it is open) that it also does not intersect any of $\overline{S_i}$'s.

Thus, the complement of $\bigcup_{i\in I}\overline{S_i}$ is open, i.e. $\bigcup_{i\in I}\overline{S_i}$ is closed.