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Thread: Closure (Metric Space)

  1. #1
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    Closure (Metric Space)

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

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

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

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

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