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Here's a question I've been working on:

Suppose that $X$ is a normal topological space, that $F\subseteq X$ is closed, and that $F\subseteq U_1 \cup U_2$ for open sets $U_1,U_2$. Prove that there exist closed sets $F_1,F_2$ with $F=F_1 \cup F_2$, $F_1 \subseteq U_1$, and $F_2 \subseteq U_2$.

I wish I could put up a partial solution, but I don't have much. I'm having a hard time seeing where I could apply normality---where are my disjoint closed sets?

Here's what I do know:

(1) closed subspaces of normal spaces are normal;

(2) $X$ is normal iff given a closed set $F$ and an open set $U$ containing $F$, there exists an open set $V$ such that $F \subseteq V\subseteq \bar{V} \subseteq U$.

I'm thinking that the characterization of normality given in (2) might be more helpful than the actual definition of normality (since it gives me closed sets contained in open sets), but again, I'm having a difficult time applying it.

Any hints would be appreciated.

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Hint: The complement of any open set (containing a particular closed set) is closed (and disjoint from the initial closed set). Also, (2) will probably be more useful, since $F$ is a subset of the open set $U_1\cup U_2$.

You might also find useful the following: if $C$ is closed and $V$ is open in a given space, then $C\smallsetminus V$ is closed and $V\smallsetminus C$ is open.

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    Yeah, minor tweaks to the proof given in the other question will get you there. They're looking for open sets and we're looking for closed sets, but the changes in the argument are small. It suffices to focus on $F$ ($F$ is normal and $F=(U_1\cap F)\cup (U_2\cap F)$ where the sets are relatively open) and forget about the ambient space since a set that is closed relative to a closed subspace is closed in the ambient space.2012-05-18