how to prove closure of unions is equal union of closures in paracompactness

Question

Can anyone show me how to prove that: In a paracompact space: given a family of sets $A_i$ where $i \in I$ (some set of indexes) then: $\overline{\cup{A_i}} = \cup \overline{A_i}$ Thanx!

Answer 1

This is true, given that all $A_i$ are open.

First off, we get $\cup \overline{A_i} \subseteq \overline{\cup{A_i}}$ due to the nature of the closure operator.

Now let $x \in \overline{\cup A_i}$. Since the space is paracompact, and since $\cup A_i$ is an open cover, $\cup A_i$has some locally finite refinement $\cup B_i$, so there is some neighbourhood $U$ of $x$ such that $U$ intersects finitely many $B_i$.

Suppose towards contradiction that $x \notin \cup \overline{A_i}$. $\cup \overline{B_i} \subseteq \cup \overline{A_i}$, so $x \notin \cup \overline{B_i}$, hence for each $B_i$, $x \notin \overline{B_i}$.

Now let $i_1, i_2, ..., i_n$ be the indices such that $U \cap B_{i_1}, U \cap B_{i_2}, ..., U \cap B_{i_n}$ are all non-empty, and let $V = U \cap (\displaystyle \cup_{k=1}^n \overline{B_{i_k}}')$ ($U$ subtract the compliments of the closures). Notice $V$ is an open neighbourhood of $x$ that does not intersect any $B_i$.

But that means $V$ does not intersect $\cup B_i$, so $V'$ is a closed set containing $\cup B_i$ such that $x \notin V$, which means $x \notin \overline{\cup B_i}$. $\cup B_i$ is a refinement of $\cup A_i$, so $x \notin \overline{\cup A_i}$, contrary to our assumption.

That means we must in fact have $x \in \cup \overline{A_i}$, giving $\overline{\cup A_i} \subseteq \cup \overline{A_i}$, which combines with our first inequality for the equality $\overline{\cup A_i} = \cup \overline{A_i}$.

Answer 2

$\newcommand{\cl}{\operatorname{cl}}$The statement is false in general, even if the sets $A_i$ are open. Let $S$ be the Sorgenfrey line, i.e, the real line topologized by taking $\big\{[a,b):a<b\big\}$ as a base. $S$ is $T_3$ and Lindelöf, so it’s paracompact. For $x\in(0,1)$ let $U_x=[x,1)$, and let $\mathscr{U}=\{U_x:x\in(0,1)\}$. $\mathscr{U}$ is a family of clopen sets in $S$, so $$\bigcup_{x\in(0,1)}\cl_SU_x=\bigcup_{x\in(0,1)}U_x=(0,1)\;,$$ but $$\cl_S\bigcup_{x\in(0,1)}U_x=\cl_S(0,1)=[0,1)\;.$$

The closest result that I can think of is this one:

Theorem. Let $X$ be a space, and let $\mathscr{A}$ be any locally finite family of subsets of $X$. Then $$\cl_X\bigcup\mathscr{A}=\bigcup\big\{\cl_XA:A\in\mathscr{A}\big\}\;.$$

Proof. Clearly it suffices to show that $\cl_X\bigcup\mathscr{A}\subseteq\bigcup\big\{\cl_XA:A\in\mathscr{A}\big\}$, so suppose that $x\in\cl_X\bigcup\mathscr{A}$. Let $V$ be an open nbhd of $x$ meeting only finitely many members of $\mathscr{A}$, and let $\mathscr{A}_0=\{A\in\mathscr{A}:V\cap A\ne\varnothing\}$. If $U$ is any open nbhd of $x$, so is $U\cap V$, so $(U\cap V)\cap\bigcup\mathscr{A}\ne\varnothing$. But $V\cap\bigcup\mathscr{A}=V\cap\bigcup\mathscr{A}_0$, so $(U\cap V)\cap\bigcup\mathscr{A}=(U\cap V)\cap\bigcup\mathscr{A}_0$ and it follows that $x\in\cl_X\bigcup\mathscr{A}_0$. But $\mathscr{A}_0$ is finite, so $\cl_X\bigcup\mathscr{A}_0=\bigcup\big\{\cl_XA:A\in\mathscr{A}_0\big\}$, and there is therefore some $A_x\in\mathscr{A}_0$ such that $x\in\cl_XA_x\subseteq\bigcup\big\{\cl_XA:A\in\mathscr{A}\big\}$. $\dashv$