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Confusion about definition of compactness

  1. Aug 16, 2015 #1
    So the definition I have seen is:

    Given a topological space <S,F> it is compact if for any cover (union of open sets which is equal to S) there exists a finite subcover.

    By the definition of a topological space both S and the empty set must belong to the family of subsets F.
    Wouldn't <S, empty set> be a finite subcover for S? In which case S is compact.

    By this sort of logic any open subset X of a topological space S is also compact in the relative topology since X will belong to the family of subsets in the relative topology so <X, empty set> would be a finite subcover for X.

    I'm assuming the resolution to this is that the finite subcover cannot include the space itself but I just want to double check I haven't horribly misunderstood something.
  2. jcsd
  3. Aug 16, 2015 #2


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    {S} is a finite cover of S. It's one which exists for every topology on S, and it proves nothing about compactness of S with respect to that topology.

    There may be other covers of S which don't have {S} as a subcover (for example, {{1},{2}} is an open cover of {1,2} in the discrete topology), and don't have any finite subcover ([itex]\{ (-n,n) : n \in \{1, 2, 3, \dots\}\}[/itex] is an open cover of [itex]\mathbb{R}[/itex] in the standard topology which does not admit a finite subcover).
  4. Aug 16, 2015 #3


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    You've effectively taken the definition of compactness to be "there exists a finite open cover".
  5. Aug 16, 2015 #4


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    ##\{\mathbb R\}## is an open cover of ##\mathbb R## that has a finite subcover. But this tells us nothing about whether ##\mathbb R## is compact. The set of all open intervals in ##\mathbb R## is an open cover of ##\mathbb R## that doesn't have a finite subcover. This tells us that ##\mathbb R## is not compact.

    The space itself can be included in the finite subcover, but only if it's also an element of the original open cover, the one that the subcover is a subset of. If you prove that every open cover that includes the space itself has a finite subcover, you have only proved a trivial statement about some open covers.

    For a moment I thought that the proof of "closed subsets of compact spaces are compact" would be a good example of an argument where you need to include the space in an open cover and its subcover, but I realized that I had made a mistake right after I had typed up the proof (incorrectly) and posted it. The corrected proof is still an interesting example of how the definition is used, so I'll include it here, even though it doesn't illustrate the point that I was originally trying to make.

    Let X be a compact topological space. Let F be a closed subset of X. Let C be an open cover of F. ##C\cup\{F^c\}## is an open cover of X. Let D be a finite subset of ##C\cup\{F^c\}## that covers X. (The fact that X is compact ensures that such a D exists). ##D-\{F^c\}## is a finite subset of C that covers F. Since C is an arbitrary open cover of F, this implies that F is compact.
    Last edited: Aug 16, 2015
  6. Aug 16, 2015 #5


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    You need to interpret "for any" to mean "every". So even a cover that does not include S must contain a finite subcover.
  7. Aug 16, 2015 #6
  8. Aug 17, 2015 #7
    Thank you for all the responses!

    My problem was that {S, empty set} is a finite cover of S since the union of S and the empty set is just S.
    Then surely {S} is a subset of {S, empty set} in which case is this not a finite subcover?

    I think the answer to this problem lies with remark FactChecker made. I think that indeed I have proven that there exists an open cover with a finite subcover however I have not proven that for EVERY open cover there exists a finite subcover and this is what is required to state that S is compact.
  9. Aug 19, 2015 #8


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    That is correct. One always has a finite cover which is just the whole space itself. Compact means as you said - every open cover has a finite subcover.
  10. Aug 20, 2015 #9


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    For example, the set of all integers, as a subset of the set of all real numbers with the usual topology, is NOT compact because the "open cover", {(n- .01, n+ .01}, has no open subcover. Yes, the empty set and the entire set of integers cover it but that is not a sub collection of this particular open cover.
    Last edited: Aug 20, 2015
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