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Open sets and closures

  1. Jan 27, 2014 #1
    Suppose A and B are open sets in a topological Hausdorff space X.Suppose A intersection B is an empty set. Can we prove that A intersection with closure of B is also empty? Is "Hausdorff" condition necessary for that?

    Please help.
  2. jcsd
  3. Jan 27, 2014 #2


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    Given that [itex]A[/itex] and [itex]B[/itex] are disjoint, the only way [itex]A[/itex] can intersect with the closure of [itex]B[/itex] is if there exists [itex]a \in A[/itex] such that [itex]a[/itex] is a limit point of [itex]B[/itex].

    But that's impossible: [itex]A[/itex] is an open neighbourhood of [itex]a[/itex] which contains no points in [itex]B[/itex]. Hence [itex]a[/itex] is not a limit point of [itex]B[/itex].

    This holds whether or not [itex]X[/itex] is Hausdorff.
  4. Jan 27, 2014 #3
    Rephrasing pasmith's argument without reference to individual points...

    Let [itex]X[/itex] be any topological space, and suppose [itex]A,B\subseteq X[/itex] are disjoint and [itex]A[/itex] is open. Then:
    -[itex]X\setminus A[/itex] is closed because [itex]A[/itex] is open.
    -[itex]X\setminus A \supseteq B[/itex] because [itex]A,B[/itex] are disjoint.
    -As the closure of [itex]B[/itex], the set [itex]\bar B[/itex] is the smallest closed set that contains [itex]B[/itex].
    -In particular, [itex]\bar B \subseteq X \setminus A[/itex].
    Rephrasing the last point, [itex]A,\bar B[/itex] are disjoint sets.
  5. Jan 28, 2014 #4
    Thank you very much
  6. Feb 4, 2014 #5
    A \cap B empty

    If A and B are disjoint, the B is a subset of the complement of A.

    If A is open, its complement is closed.

    Hence, in this case, the closure of B is contained in the complement of A.

    Hence, A and the closure of B are disjoint.

    There is no need for the ambient space to be Hausdorff. There is no need for B to be ooen.
  7. Feb 5, 2014 #6
    The space has to be Hausdorff and B has to be open. A = (0,1) and B = [1,0] is a counter example to what you say
  8. Feb 5, 2014 #7
    1. [1,0] is the empty set, as there are no real numbers which are both at least 1 and at most 0. Hence your "counter-example" fails.

    2. By definition,
    (i) a subset of a topological space is closed if and only if it is the complement of an open set
    (ii) the closure of a subset, B, of a topological space is the smallest closed subset of the space which contains B.

    There is no question of being a Hausdorff space or even a T1 or T0 space for this.
  9. Feb 5, 2014 #8
    Wow, I can't believe I wrote that. I was dead tired and my counter-example I was trying to type out was A = (0,1), B = [1,2]. But even that fails.

    My apologies, I'll try not to post when I'm half awake anymore lol.
  10. Feb 5, 2014 #9
    I'm glad I'm not the only one to blunder at times!
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