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Math Help - Closure (Metric Space)

  1. #1
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    Closure (Metric Space)

    Let A, B be subsets of a metric space. Show that  \overline{A \cup B} =  \bar{A} \cup \bar{B} and that  \overline {A \cap B} \subset \bar{A} \cap \bar{B} . Give an example to show that  \overline {A \cap B} and  \bar{A} \cap \bar{B} may not be equal.
    Last edited by aliceinwonderland; January 11th 2009 at 08:08 PM.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by aliceinwonderland View Post
    Let A, B be subsets of a metric space. Show that  \overline{A \cup B} = \bar{A} \cup \bar{B} and that  \overline {A \cap B} \subset \bar{A} \cap \bar{B} .
    What are you having trouble with?
    Give an example to show that  \overline {A \cap B} and  \bar{A} \cap \bar{B} may not be equal.
    Consider when A,B are seperated but share a limit point. Example A=(-1,0),B=(0,1) because A\cap B=\varnothing\implies\overline{A\cap B}=\varnothing but 0 is a limit point of both so \bar{A}\cap\bar{B}=\left\{0\right\}
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  3. #3
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    Quote Originally Posted by aliceinwonderland View Post
    Let A, B be subsets of a metric space. Show that  \overline{A \cup B} = \overline{A} \cup \overline{B} and that  \overline {A \cap B} \subset \overline{A} \cap \overline{B} . Give an example to show that  \overline{A \cap B} and  \overline{A} \cap \overline{B} may not be equal.

    a) One inclusion: \overline{A} \cup \overline{B} is closed since it is the finite union of closed sets and it contains A \cup B so it must contain the closure \overline {A \cup B}.

    Other inclusion: STS \cup \overline{A_\alpha} \subseteq \overline{\cup A_\alpha}. Well, \forall \alpha, A_\alpha \subseteq \overline{\cup A_\alpha}. So \overline{A_\alpha} \subseteq \overline{\cup A_\alpha}. Therefore \cup \overline{A_\alpha} \subseteq \overline{\cup A_\alpha}.

    Next Part: A \cap B is closed set containing A \cap B. So it must contain its closure \overline{A \cap B}

    next part: take A=(0,1) and B=(-1,0) then \overline{A \cap B} = \emptyset but \bar{A} \cap \bar{B}=\{0\}. QED
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