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Compact sets definition 
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#1
Feb2312, 09:52 PM

P: 252

A subset K of a metric space X is said to be compact if every open cover of K contains a finite subcover.
Does not this imply that every open set is compact. Because let F is open, then F= F [itex]\bigcup[/itex] ∅. Since F and ∅ are open , we obtained a finite subcover of F. Am I missing something here? 


#2
Feb2312, 09:58 PM

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#4
Feb2312, 10:09 PM

P: 800

Compact sets definition
Compactness says that if you have an open cover  a collection of open sets whose union covers F  then some finite subcollection will also cover F. In other words you have to be able to pick out a finite collection of sets from your open cover, and show that the finite collection is also an open cover. As an example, take F = (0,1), the open unit interval. Let A_n = (1/n, 1). Then the collection {A_n} for n = 1, 2, 3, ... is an open cover of F; but no finite subcollection of the A_n's covers (0,1). Working through this example will be helpful. 


#5
Feb2312, 10:15 PM

P: 252

Then, I think the definition of the open cover of a set F has to emphasize that F has to be a proper subset of the cover. Otherwise the argument in my first post will be correct.



#6
Feb2312, 10:17 PM

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#7
Feb2312, 10:19 PM

P: 800

But compactness requires that EVERY open cover has a finite subcover. I think you would find it helpful to work through the example of F = (0,1) that I gave earlier. FWIW pretty much everyone has trouble learning about compactness. It's very unintuitive. Not like connectedness, say, whose technical definition matches our intuition about what a connected set should be. With compactness, it's like, "How did they ever think of that?" 


#8
Feb2312, 10:23 PM

P: 252

SteveL27 and lavinia thank you very much for your help! I have a better understanding now of the definition. The open unit interval example was particularly helpful.



#9
Feb2412, 01:00 AM

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P: 906

the usual example "in the other direction" (unbounded sets, rather than open sets) is R:
certainly {R} is an open cover of R. yet it hardly seems "right" to call R "compact", given how large it is. and, of course, it is not: {(n,n+2): n in Z} is an open cover of R, but we need "all Z of them" (and Z is not finite) to cover R, no finite subcover will do. 


#10
Feb2412, 07:30 AM

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P: 1,716

Take the integers with the discrete topology. Here each singleton,{n}, is an open set. Cover the integers with the set of singletons. Not onl;y is there no finite subcover, there is no subcover. 


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