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Compact Sets

  • Thread starter glebovg
  • Start date
  • #1
164
1
Homework Statement

Show that every compact set must be closed.

I am looking for a simple proof.

This is supposed to be Intro Analysis proof.

Relevant equations

Any compact set must be bounded.

The Attempt at a Solution



Suppose A is not closed, so let a be an accumulation point of A such that a is in A etc. (I think I can finish this proof, but I do not like it.)

Does anyone know of a simpler proof.
 

Answers and Replies

  • #2
662
1
I don't understand: if you are working in R^n (which I imagin you may be doing, since you mention an analysis book), then Heine-Borel is the resul. If not, the result is true only if you have a compact subset of a Hausdorff space, where you show that the complement in the space of the compact set is open, by finding a 'hood (neighborhood) of a point x in the complement that is disjoint from the compact space, by using Hausdorffness (so you can choose disjoint pairs of open sets containing x, disjoint from finitely-many open subsets of the cover. This is the main idea ). So this last one should take care of the case of R^n, or for any Hausdorff space.
 
  • #3
22,097
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Yes, what is the context you are working in?? Are you working in [itex]\mathbb{R}^n[/itex]?? In a metric space? A topological space? A Hausdorff space??

And how did you define compact??
 
  • #4
164
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The section in the book is called Topology of the Real Number System. We have not covered metric spaces, topological spaces, Hausdorff space, etc.

A set A subset R is compact if every open cover C of A contains a finite subcover F of A.

And by the way can anyone explain what does a cover, a subcover, etc. really represent? I understand the definitions but what is really meant by something like C covers A?
 
  • #5
662
1
C covers A if the union of the open sets contains A; a subcover of C is a subcollection of the subsets of C, so you want a finite subcollection that contains C.

An example, maybe somewhat-trivial, is that of a convergent sequence {xn} in R, with limit x; take a collection of open intervals (an,bn)--take it, please! (sorry, prepping for my 2nd job)-- with an:= xn-0.1, (any small number will do ), and bn:=xn+0.1, and a 'hood (neighborhood) (x-0.1,x+0.1) of the limit point x . Then the union of the intervals (a,n,bn) and (x-0.1,x+0.1)covers {xn}, together with a 'hood , trivially, since each member of {xn} is in the n-th interval. Now, you can find a subcover ( you, and you alone!), by using (x-0.1,x+0.1) , which , by convergence of {xn}, will contain all-but-finitely many of the elements of the sequence, so that you can use finitely-many intervals (an,bn) to cover all the remaining elements.
 
  • #6
662
1
It may be good too, for contrast, to consider a cover of the integers in the real line, by
open intervals (n-1/4,n+1/4), and see if you can find a finite subcollection of these open sets that covers the integers. For a more advanced problem, show that a half-open interval [a,b) in the real line is not compact. I'm pretty sure you will eventually see Heine-Borel theorem which tells you that, for Rn, the compact sets are always those that are close and bounded.
 
  • #7
HallsofIvy
Science Advisor
Homework Helper
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If you are working in a general vector space or general metric space, use the fact that a set is closed if its complement is open. To show that the complement of a set is open, show that any point NOT in the set (and so in the original compact set) has an open neighborhood that does not intersect the intended closed set.
 

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