The union of any collection of closed sets is closed?

rhino1000
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I don't see how this is the case. Let ao and bo be members of [A,B] with ao<bo. Let {ai} be a strictly decreasing sequence, with each ai>A and {bi} be a strictly increasing sequencing with each bi<B. Let the limits of the two sequences be A and B, respectively. Then define Ii = [ai,bi]. It seems to me that the union of {Ii} is an open set, not a closed set. Thoughts?
 
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For others wondering, It can be shown that, in any topological space, the union of any collection of open sets is open. and the intersection of any finite collection of open sets is open. On the other hand, the intersection of any collection of closed sets is closed and the union of any finite collection of closed sets is closed.

It is not true, in general, that the intersection of any collection open sets is open or that the union of any collection of closed sets is closed. Counter-examples: all the sets in the collection {(1/n, (n-1)/n)} are open but the intersection of all sets in that collection is the singleton set, {1/2}, which is not open. On the other hand, all the sets in the collection {[1/n, (n-1)/n]} are closed but the union of all sets in that collection is (0, 1) which is not closed.
 
picky comment: in these last nice examples, it is necessary to state what the topological space is under consideration, since open and closed are relative terms. of course it is apparent that the space of real numbers is meant. E.g. in the last example, should the space have been taken as (0,1), the statement that (0,1) is not closed would of course, not have been true.
 
rhino1000 said:
I don't see how this is the case. Let ao and bo be members of [A,B] with ao<bo. Let {ai} be a strictly decreasing sequence, with each ai>A and {bi} be a strictly increasing sequencing with each bi<B. Let the limits of the two sequences be A and B, respectively. Then define Ii = [ai,bi]. It seems to me that the union of {Ii} is an open set, not a closed set. Thoughts?
The union of an arbitrary collection of closed sets is not in general closed. One of the axioms of topology says that a finite union of closed sets is closed, however it can be proved that a family of closed sets needs only be locally finite for the union to be closed.
 
Just take an intersection of opens that is not open -- standard is that the intersection is a point, and then use DeMorgan to find a counter for
unions of closed sets.
 
Cruz Martinez said:
The union of an arbitrary collection of closed sets is not in general closed. One of the axioms of topology says that a finite union of closed sets is closed, however it can be proved that a family of closed sets needs only be locally finite for the union to be closed.

An interesting question is to characterize those spaces where the arbitrary union of closed sets is closed. Those spaces are called "finitely generated spaces" or "Alexandroff spaces". They are characterized completely by order theory. They are surprisingly useful too!
 
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Any interesting Alexandroff spaces other than discrete or finite spaces?
 
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micromass said:
An interesting question is to characterize those spaces where the arbitrary union of closed sets is closed. Those spaces are called "finitely generated spaces" or "Alexandroff spaces". They are characterized completely by order theory. They are surprisingly useful too!
The dual notions of ##G_{\delta}## and ##F_{\sigma}## sets may also be of interest in this context. One can then inquire about spaces where merely every ##F_{\sigma}## is closed.

EDIT: Found this: https://en.wikipedia.org/wiki/P-space which also answers
WWGD said:
Any interesting Alexandroff spaces other than discrete or finite spaces?
Amusing material.
 
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