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Compact sets definition

  1. Feb 23, 2012 #1
    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. jcsd
  3. Feb 23, 2012 #2


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    Every cover must have a finite subcover not just one or some of them.
  4. Feb 23, 2012 #3
    But F[itex]\bigcup[/itex]∅ is contained in every open cover for F?
  5. Feb 23, 2012 #4
    As a subset, yes. But F and ∅ might not be elements of the open cover.

    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.
  6. Feb 23, 2012 #5
    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.
  7. Feb 23, 2012 #6


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    As SteveL7 said, F and ∅ may not be elements of the open cover. An open cover is a collection of open sets. F may not be in the collection.
  8. Feb 23, 2012 #7
    Well if F is an open set, then {F} is indeed an open cover of F consisting of one set. So thats an example of one open cover of F that happens to have a finite subcover.

    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?"
  9. Feb 23, 2012 #8
    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.
  10. Feb 24, 2012 #9


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    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.
  11. Feb 24, 2012 #10


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    Here is a stark example.

    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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