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Math Help - Metric spaces and closed sets

  1. #1
    MHF Contributor Siron's Avatar
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    Metric spaces and closed sets

    Hi, I need some help with the following question:

    Let (X,d) be a metric space and A \subseteq X. Define the set A^{(c)} \subseteq X for \epsilon>0 as \{ x \in X | d(x,A) \leq \epsilon \}

    Prove the following statements:
    (1)  \forall \epsilon \geq 0: \overline{A} \subseteq A^{(c)}
    (2)  \overline{A} = \bigcap_{\epsilon \geq 0} A^{(c)}
    (3)  A^{(c)} \ \mbox{is closed}
    (4)  \partial A^{(c)} \subseteq \{ x \in X | d(x,A)=\epsilon \}
    (5) Does there exist an example wherefore \partial A^{(c)} \subset \{ x \in X | d(x,A)=\epsilon \}

    Proofs:
    (1) Suppose  x \in \overline{A} thus by definition d(x,A)=0 \leq \epsilon (with \epsilon \geq 0) therefore x \in A^{(c)}

    Can someone give some hints to prove the other 4 statements, because I'm stuck there.
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    Re: Metric spaces and closed sets

    Quote Originally Posted by Siron View Post
    Hi, I need some help with the following question:

    Let (X,d) be a metric space and A \subseteq X. Define the set A^{(c)} \subseteq X for \epsilon>0 as \{ x \in X | d(x,A) \leq \epsilon \}

    Prove the following statements:
    (1)  \forall \epsilon \geq 0: \overline{A} \subseteq A^{(c)}
    (2)  \overline{A} = \bigcap_{\epsilon \geq 0} A^{(c)}
    (3)  A^{(c)} \ \mbox{is closed}
    (4)  \partial A^{(c)} \subseteq \{ x \in X | d(x,A)=\epsilon \}
    (5) Does there exist an example wherefore \partial A^{(c)} \subset \{ x \in X | d(x,A)=\epsilon \}

    Proofs:
    (1) Suppose  x \in \overline{A} thus by definition d(x,A)=0 \leq \epsilon (with \epsilon \geq 0) therefore x \in A^{(c)}
    From #1 you already have half of #2. So show that  \overline{A} \supset \bigcap_{\epsilon \geq 0} A^{(c)}.

    In #3 I would show the complement is open.

    #4 should fall right from applying the definition of boundary point.
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  3. #3
    MHF Contributor Siron's Avatar
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    Re: Metric spaces and closed sets

    To prove: \bigcap_{\epsilon \geq 0} A^{\epsilon} \subset \overline{A}
    Proof:
    Suppose x \in \bigcap_{\epsilon\geq 0} A^{\epsilon} = A^{\epsilon_1} \cap A^{\epsilon_2} \cap \ldots
    thus  x \in A^{\epsilon_1} \wedge x \in A^{\epsilon_2} \wedge \ldots which means
     d(x,A) \leq \epsilon_1 \wedge d(x,A) \leq \epsilon_2 \wegde \ldots and therefore
     d(x,A) \leq \max\{\epsilon_1,\epsilon_2,\ldots\}, write d(x,A)\leq e'
    But d(x,A)=\inf\{d(x,a)|a \in A \} \leq e', therefore we can choose an element  a \in A with d(x,a) \leq \epsilon' \wedge a \in A
    and so \forall \epsilon' \geq 0: B(x,\epsilon') \cap A \neq \emptyset \Rightarrow x \in \overline{A}

    What do you think about this proof?
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    Re: Metric spaces and closed sets

    Quote Originally Posted by Siron View Post
    To prove: \bigcap_{\epsilon \geq 0} A^{\epsilon} \subset \overline{A}
    What would it mean if \exists y\in\left( \bigcap_{\epsilon \geq 0} A^{\epsilon}\right) \setminus \overline{A}.

    Let \delta=D(A;y)>0. If c=\tfrac{\delta}{2} can y\in A^{(c)}~?
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  5. #5
    MHF Contributor Siron's Avatar
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    Re: Metric spaces and closed sets

    Quote Originally Posted by Plato View Post
    What would it mean if \exists y\in\left( \bigcap_{\epsilon \geq 0} A^{\epsilon}\right) \setminus \overline{A}.
    It means that \exists y \in \bigcap_{\epsilon \geq 0} A^{\epsilon} \wedge y \notin \overline{A}

    Quote Originally Posted by Plato View Post
    Let \delta=D(A;y)>0. If c=\tfrac{\delta}{2} can y\in A^{(c)}~?
    I never have seen something like \delta=D(A;y) before, what does it mean?
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    Re: Metric spaces and closed sets

    Quote Originally Posted by Siron View Post
    \delta=D(A;y) before, what does it mean?
    \delta=D(A;y) denotes ther distance from y to A.
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