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Intersection of open sets

  1. Jul 29, 2007 #1
    1. The problem statement, all variables and given/known data

    Prove that any set in a metric space is an intersection of open sets.

    2. Relevant equations



    3. The attempt at a solution

    I think the general idea would be that this set is in a (probably infinite) number of open sets, so we just take the smallest one. But I am not sure how to rigorize that...
     
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  3. Jul 29, 2007 #2

    CompuChip

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    I think the rigorous argument is: X is contained in the intersection of all open sets that contain X.
    Which does give you the smallest open set you spoke about. The problem is of course to prove that taking this intersection is possible (is it not empty, for example? why should there be any set in the intersection at all?)
     
  4. Jul 29, 2007 #3
    I think we can assume the set is non-empty. Even if it is empty, it would then be the intersection of any two disjoint sets or the empty set and any other set. I think we must assume that the metric space is non-empty.

    So, assume the X is non-empty. We have that it is in the intersection of all open sets that contain X. It is not difficult to prove that an intersection of open sets is open, so we must only prove that there are no points P outside of X in the intersection of all open sets that contain X.

    However, P cannot be in the intersection, because the complement of P is an open set that contains X.

    How's that?
     
  5. Jul 29, 2007 #4

    Dick

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    An infinite intersection of open sets is not necessarily open. Intersect (-1/n,1/n). You are going to have to be more specific about some of your open sets. Hint: this is a metric space. If x is a point in X, what kind of set is the complement of {x}?
     
    Last edited: Jul 29, 2007
  6. Jul 29, 2007 #5

    CompuChip

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    Which is good, right?
    I mean, the question is to prove that any set can be written as an intersection of opens. If any intersection of opens were open, then the theorem wouldn't hold for a lot of sets, in particular all closed ones (except the empty set and the entire space).
     
  7. Jul 29, 2007 #6
    What is n? If it is in R, the intersection is {R} - 0, which is open.


    You're right--we do not even need to show that X is an open set. Let me try again.

    Let Y be the intersection of all open sets that contain X. We will show that X = Y. So, X is in Y. We must show that Y is in X. If there is a point P in Y but not X, then the complement of point P in a set Z that contains X must be open (in fact the complement of point P in all sets that contain X must be open).

    This complement is now an open set that contains X but not P leading to a contradiction of our assumption that Y is the intersection of all open sets that contain X.
     
  8. Jul 29, 2007 #7

    CompuChip

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    He meant, what is
    [tex]\bigcap_{n \in \mathbf{N}} ]-1/n, 1/n[ [/tex]

    Looks good. I think for complete rigor you still have to remark that the entire space is open and contains X, and therefore the intersection Y contains at least the entire space and hence is non-empty. And you probably want to quote some lemma or theorem to support the first paragraph of the above quote (about the complement of a point), that uses the metrisability.

    But else, it seems fine
     
  9. Jul 29, 2007 #8
    So, you mean the intersection:

    [tex](-1,1) - 0 \cap (-1/2,1/2) -0 \cap (-1/3, 1/3) - 0 \cap ... [/tex]

    which is a smaller and smaller interval around zero?


    Why is that intersection not open since those are open intervals?
     
    Last edited: Jul 29, 2007
  10. Jul 29, 2007 #9

    morphism

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    That intersection is the singleton {0}, which isn't open.

    We know that the intersection of finitely many open sets is open. This example should show you that we can't even loosen "finite" to "countable".
     
  11. Jul 30, 2007 #10

    CompuChip

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    The second sentence, yes. But the first sentence... why did you take out the zero? It's just
    [tex](-1,1) \cap (-1/2,1/2) \cap (-1/3, 1/3) \cap \cdots [/tex]

    Then the only point that all sets in the intersection really have in common is zero (otherwise it would be empty). If you like, you can try proving this (suppose there is a point other than zero in the intersection. Use that it's a metric space and/or the rationals are dense).
     
  12. Jul 30, 2007 #11
    You're right. I should not have taken out 0.
     
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