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Nested set property in a general topological space

  1. Jul 1, 2007 #1

    quasar987

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    1. The problem statement, all variables and given/known data
    The nested set property: "If {F_k} is a decreasing sequence of non empty compact sets in a metric space (M,d), then their intersection is non empty."

    First I cooked up a proof of my own and then I read the one provided by the book. It seemed to me that they were perfectly applicable to a general topological space... the metric structure did not seem to come in. But I soon realized that they both heavily relied on the fact that compact ==> closed, which, is not necessarily true in a topological space (Any proper subset of A is compact under the discrete topology but none is closed).

    So I was just wondering if in the context of topological spaces, the nested set property is a feature unique to metrizable spaces, or is there a class of conditions more broad that metrizability that make the nested set property true?
     
  2. jcsd
  3. Jul 1, 2007 #2

    StatusX

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    The proof I'm familiar with uses the metric in a pretty essential way. Do you mind sketching the proof you have in mind that doesn't refer to the metric?
     
  4. Jul 1, 2007 #3

    HallsofIvy

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    More general is the "finite intersection property": If a family of compact sets has the property that the intersection of any finite number is non-empty, then the intersection of all of them is non-empty.
     
  5. Jul 1, 2007 #4

    quasar987

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    Halls: Is this true in a general topological space or only in a metric space?

    I know the finite intersection property (F.I.P.) as something substantially different: A collection of closed sets in a topological space X is said to have the F.I.P. with a subset A if the intersection of any finite number of these closed sets with A is non-empty.

    Thm: (Caracterisation of compactness) A set A in a topological space X is compact iff whenever A has the F.I.P. with a collection of closed sets, the intersection of A with the whole family is non empty.

    (This can be thought of as just an alternative topological definition of compactness in terms of closed sets, rather than open sets).

    From there, your "finite intersection property" follows trivially, but in a metric space only because in there, compact ==> closed.

    StatusX: I'm killing two birds with one stone here because the easiest proof of the nested set property is a direct corollary of the caracterisation stated above, provided we have the following:

    Lemma: Compact sets in metric spaces are closed. I.e. sequentially compact sets are closed.

    Proof of N.S.P.: F_1 is compact and {F_k} is a family of closed sets that has the F.I.P with F_1. Therefor, the intersection of the lot is non-empty.

    Sketch of another proof w/o the F.I.P.: Suppose the intersection is void, then by DeMorgan, this is equivalent to say that the union of their complements is the whole of M. In particular, it means that {M\F_k} is an open cover of F_1. Since F_1 is compact, we should be able to find a finite subcover, but it is clear "geometrically" that we can't.
     
  6. Jul 2, 2007 #5

    morphism

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    The common definition for the FIP is what you've given but applied to the set A=X. Then in a compact space X, the nested set property you described in your first post holds, as a corollary of the theorem you just posted.

    I don't think there's any other generalization.
     
    Last edited: Jul 2, 2007
  7. Jul 19, 2007 #6

    quasar987

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    Not a generalization but here's a variant:

    Since in a metric space, compact is equivalent to closed + tot. bounded, the nested set property holds if you trade the tot. bounded property for completeness of the metric space and add the condition that the diameter of each sets in the collection goes to 0.

    In other words, a collection of decreasing closed sets {F_k} in a complete metric space such that diam(F_k)-->0 has a non-empty intersection.

    (This is because closed sets in a complete metric space are complete as subspaces. So if you define a sequence by taking x_n in F_n, then the sequence is Cauchy because diam(F_k)-->0 and thus converges to a point that belongs to each F_k, thus that limit is in the intersection. In fact, diam(F_k)-->0 also implies that the intersection consists of this only point.)
     
  8. Jul 19, 2007 #7

    matt grime

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    Try looking up Noetherian topological space.
     
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