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There is the nice theorem that the union of a locally finite collection of closed sets is closed. Are there known any other natural conditions on a collection of closed sets which imply this?

There are collections of closed sets with closed unions which are not themselves locally finite. For example, take the singleton subsets of $\mathbb{R}$. Therefore "locally finite" cannot be the best possible condition in the sense that any other condition implying the closed union condition implies it.

Geoffrey Sangston
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    Michael's theorem may interest you. – Ittay Weiss May 25 '17 at 20:00
  • Are you asserting that the union of any collection of singleton subsets of $(0,1)$ is again closed? That's definitely false. – Greg Martin May 25 '17 at 20:00
  • Sorry Greg I was misremembering an earlier thought with that example. The revised question should hopefully make things clear. Basically not all collections with closed unions have to be locally finite, so there's no "if and only if" theorem here. This is the sense that locally finite might not be best possible. – Geoffrey Sangston May 25 '17 at 20:29

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I can propose to you the following simple characterization.

Let $\mathcal F$ be a family of closed subsets of a space $X$. Then $F=\bigcup\mathcal F$ is closed in $X$ iff $\mathcal F$ is locally finite at each point $x\in X\setminus F$, that means that a point $x$ has a neighborhood intersecting only finitely many members of the family $\mathcal F$. Indeed, if $F$ is closed then each point $x\in X\setminus F$ has a neighborhood ($X\setminus F$) intersecting no members of the family $\mathcal F$. Conversely, if a point $x$ has a neighborhood $O_x$ intersecting only finitely many members of the family $\mathcal F$ then $O_x\setminus\bigcup \{S\in\mathcal F:S\cap O_x\ne\varnothing\}$ is an open neighborhood of the point $x$ disjoint from $F$.

Alex Ravsky
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    Hi @Alex.

    This is a great answer. I'll never assume a nontrivial iff condition doesn't exist again. Is this statement listed as an exercise or lemma in a textbook somewhere or did you just come up with it? I'm reworking through Munkres and don't see it anywhere which I find surprising given how nice it is.

     – Geoffrey Sangston May 26 '17 at 02:26
  • Hi, @Geoffrey. Thanks for your kind words. Yes, I just came up with the statement guided by my intuition of a professional topologist. As Nicholas Bourbaki wrote “the mathematician does not work like a machine, nor as the workingman on a moving belt; we can not over-emphasize the fundamental role played in his research by a special intuition, which is not the popular sense-intuition, but rather a kind of direct divination (ahead of all reasoning) of the normal behavior, ... – Alex Ravsky May 26 '17 at 04:31
  • ...which he seems to have the right to expect of mathematical beings, with whom a long acquintance has made him as familiar as with the beings of the real world”. :-) By the way, tagging your question by category theory was a bad idea, because this is one of my ignored tags, so I found your question only by accident, as a blink with a relevant name while coming to the page with the newest questions in general topology. :-) – Alex Ravsky May 26 '17 at 04:31
  • Ah I did not know that tags could be ignored. I've edited my post and removed those tags now that the answer makes the question clear. – Geoffrey Sangston May 26 '17 at 05:24