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Open/closed set and interior point problem

  1. Mar 23, 2010 #1
    1. The problem statement, all variables and given/known data

    Let [tex](X,d)[/tex] be a metric space and let [tex]A \subseteq X[/tex]. Denote the interior of [tex]A[/tex] by [tex]A^o[/tex].

    2. Relevant equations

    Prove that if [tex]A[/tex] is open or closed, then [tex](\partial A)^o = \varnothing[/tex]. (Is this still true if [tex]A[/tex] is not open or closed?)

    3. The attempt at a solution

    I don't even know what [tex]\partial A[/tex] means. Can anyone tell me what this represents?
     
  2. jcsd
  3. Mar 23, 2010 #2
  4. Mar 24, 2010 #3
    [tex]
    \begin{align*}
    \partial A = \overline{A} \cap \overline{X - A} = \{ x \in
    X | \forall \varepsilon > 0 (O_\varepsilon(x) \cap A \ne \varnothing
    \wedge O_\varepsilon(x) \cap (X-A) \ne \varnothing \}
    \end{align*}
    [/tex]

    If [tex]A[/tex] is closed, then [tex]\partial A \subseteq A[/tex]. However, [tex]\forall x
    \in \partial A \forall \varepsilon > 0 ( O_\varepsilon(x) \cap (X -
    A) \ne \varnothing)[/tex], hence [tex]\forall x \in
    \partial A \nexists \varepsilon > 0 (O_\varepsilon(x) \subseteq
    \partial A)[/tex]. There are no interior points.

    If [tex]A[/tex] is open, then [tex]\partial A \subseteq X - A[/tex]. However, [tex]\forall
    x \in \partial A \forall \varepsilon > 0 ( O_\varepsilon(x) \cap A
    \ne \varnothing)[/tex], hence [tex]\forall x \in
    \partial A \nexists \varepsilon > 0 (O_\varepsilon(x) \subseteq
    \partial A)[/tex]. There are no interior points.

    Is this correct? Is it still true if [tex]A[/tex] is not open or closed? How can I prove that?
     
  5. Mar 24, 2010 #4

    jbunniii

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    Consider [itex]A = \mathbb{Q}[/itex], [itex]X = \mathbb{R}[/itex]. What is the boundary of [itex]\mathbb{Q}[/itex]?
     
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