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If I want to show a topological subspace is closed in an ambient space, does it suffice to know what happens on an open cover of the ambient space? More specifically,

Let $X$ is a topological space with a given open cover ${ U_i }$. Suppose that $Z \subset X$ is a set such that $Z \cap U_i$ is closed in $U_i$ for all $i$. Does it follow that $Z$ is closed in X?

This is clearly true if there are finitely many ${ U_i }$. At first thought, it seems unlikely to be true in the infinite case, but I'm having trouble coming up with a suitable counter-example.

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2 Answers 2

up vote 9 down vote accepted

Yes, it's true: being closed is a local property local on the base. Indeed, suppose $x \notin Z$. Then $x$ is in a neighborhood $U_i$ which is one of the sets in the open cover. Now $x \in U_i - Z \cap U_i$, which is an open set, so there is a neighborhood of $x$, contained in $U_i$, which does not intersect $Z \cap U_i$. This means that this neighborhood of $x$ does not intersect $Z$.

This is not true for closed covers (i.e. with the $U_i$ closed) in general, but it is true when they form a locally finite cover.

Incidentally, if each point $x \in Z$ has a neighborhood $U$ such that $Z \cap U$ is closed in $U$, then $Z$ is called locally closed: this means it is the intersection of a closed subset and an open subset, but is not itself necessarily closed.

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4  
+1: this is correct. Note though that, because of what you say in the last paragraph, I would say that being closed is not a local property. In other words, as a matter of terminology I disagree with the OP on the meaning of "local property". –  Pete L. Clark Aug 23 '10 at 18:15
    
Pete: Thinking about it a little more, I agree with you. It seems that a property P being local should be something like the statement "each point x has a nbd with property P implies X has property P" being true. Applied to closed sets, we get the distinction between locally closed and closed that Akhil pointed out, and so I agree that being closed is not a local property. –  Dalron Aug 23 '10 at 19:09
    
I suppose we might say, following Weyl: a property of a set is local if it can be detected by arbitrarily near-sighted policemen stationed within the set. Thinking about, say, $(0,1]$ as a subset of $[0,1]$, if the policeman at point $x \in (0,1]$ can only see a distance $x/2$, their investigation will never reveal the missing point at 0, and they cannot prove that $(0,1]$ is not closed. –  Nate Eldredge Aug 23 '10 at 19:31
    
I also agree with the comments here, and have modified the language used in the answer. –  Akhil Mathew Aug 23 '10 at 19:47
    
@Nate Eldredge But every subset is closed in its induced topology. I.e., being closed isn't an intrinsic property of a subset. Once you grant that closed is necessarily "ambient," then it is local as well, as positioning a policeman at 0 shows, in your example. I disagree with each of the above. Closed is local because when it fails it fails in every neighborhood of a point (obviously one not in the set). –  Andrew Marshall Dec 28 '10 at 17:05

$U_i \setminus (Z \cap U_i)$ is open in $U_i$, thus also open in $X$. Then $X \setminus Z = \cup_ {i \in I} U_i \setminus (Z \cap U_i)$ is open in $X$, i.e. $Z$ is closed in $X$.

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