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Thread: 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.

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

    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.

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

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

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

    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 $\displaystyle B((1,1),1)\cup B((\frac{1}{2 },-\frac{1}{2 } ),3/4)$ and $\displaystyle B((-1,1),1)\cup B((-\frac{1}{2 },-\frac{1}{2 } ),3/4)$

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

    Is there a better example for this?
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  2. #2
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    Let $\displaystyle A = (0,1)\;\& \,B = (1,2)$.
    Then $\displaystyle \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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