If compactness depends on topology, what are good examples for "alternative" topologies of spaces which steal away their compactness, i.e. I ask for spaces which are usually considered together with a topology such that they are compact and then topologies in which they are not.
Example for changing the compactness of a manifold by considering another topology
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2That is **not** the definition of compactness. (And every space has that property, since $X$ itself is in the topology of $X$.) – 2012-06-06
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0@BrianM.Scott: Okay, I was sloppy. I guess the question is sensible still? – 2012-06-06
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1I can't make any sense of this question at all. Does "spaces which are usually considered together with their usual topology such that they are compact" just mean "compact spaces"? What does this have to do with "steal[ing] away their compactness"? How is the title relevant? – 2012-06-06
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0@ChrisEagle: Mhm, I don't understand what you don't understand. But I'm not a mathematican. The answer posted basically gave a good example to me, I'll accept it soon. – 2012-06-06
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0You still haven’t stated the definition of compactness correctly. **Every** space is a finite union of open sets, because the whole space is an open set. A space $X$ is compact if every open cover of $X$ has a finite subcover. That is, whenever $\mathscr{U}$ is a collection of open sets such that $X=\bigcup\mathscr{U}$, then $X$ is the union of some finite subcollection of $\mathscr{U}$. This is completely different from (and much stronger than) saying that $X$ is the union of some finite collection of open sets. – 2012-06-06
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1@Nick: for mathematicians the word "space" _includes the topology_ as part of the data that specifies a (topological) space. If you change the topology, you _change the space._ So what you are really asking is for examples of compact topological spaces and non-compact topological spaces which have the same cardinality. – 2012-06-06
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1I think a good example occurs in functional analysis: in the dual space of an infinite-dimensional Banach space, the closed unit ball is not compact in the (usual) operator norm topology, but when you give that unit ball the weak-star topology it becomes compact. This is the Banach-Alaoglu theorem. – 2012-06-07
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Any topology on a finite set is compact, so you won't find any examples there.
However, given any infinite topological space, replacing the topology with the discrete topology always makes it non-compact.
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1Sure, but this doesn't actually have any applications. – 2012-06-07