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Math Help - Closure of intersection is a subset of intersction of closures

  1. #1
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    Closure of intersection is a subset of intersction of closures

    I just want to check that this proof is correct. This is from a section on neighbourhoods.

    \overline{(A\cap B)}\subseteq \bar{A}\cap \bar{B}

    It suffices to prove that the set of limit points of the intersection is a subset of the intersection of the set of limit points.

    (A\cap B)' \subseteq A' \cap B'

    Let p be a limit point of (A\cap B). Then there exist neighbourhoods in A or B that contain that point and other points. In fact every neighbourhood in (A\cap B) does. This means that the above statement is correct. From there we may compute

    (A\cap B) \cup (A\cap B)' \subseteq (A\cap B) \cup  A' \cap B'
    \overline{(A\cap B)}\subseteq \bar{A}\cap \bar{B}

    As required.

    Is this correct?

    There's also a second question where you have to prove that there is not equality and to come up with an example. My example is kind of inelegant and I was wondering if someone could show me another.

    Mine is this.

    Take two sets that are the union of two circles each B((1,1),1)\cup B((\frac{1}{2 },-\frac{1}{2 }  ),3/4) and B((-1,1),1)\cup B((-\frac{1}{2 },-\frac{1}{2 }  ),3/4)

    The two larger circles share a limit point at (0,1) which is outside of the intersection.

    Is there a better example for this?
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  2. #2
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    Let A = (0,1)\;\& \,B = (1,2).
    Then \overline A  \cap \overline B  = \{ 1\} \;\& \;\overline {A \cap B}  = \emptyset
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  3. #3
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    Ahh that's much more simple. Thank you.

    In retrospect that means that I could have just used the two large circles. Oh well.
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