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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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up vote 2 down vote accepted

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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This question is giving me a real hard time---could you elaborate on your hints a bit more? I'm playing with your hints, but I must be missing the point. – John Myers May 17 '12 at 17:28
When we observe (1), this question is essentially a duplicate of…. I apologize, I thought I searched the site well enough before posting. – John Myers May 17 '12 at 18:50
Not a problem--I've had that happen to me, too. I'm not so sure it is as closely related as you think, actually, but maybe I'm just too tired to see it. You don't have to accept my answer if it doesn't help you. Honestly, in looking at the answer now, I'm not sure what I was originally thinking you could do with that, so I can't really elucidate further, either. Sorry! I will take another look at it when I'm fresher, and perhaps recall what I was thinking (or come up with a better alternative. – Cameron Buie May 18 '12 at 2:15
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. – John Myers May 18 '12 at 6:16

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