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Math Help - A simple topology question

  1. #1
    Senior Member bkarpuz's Avatar
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    Exclamation [SOLVED: a counter example is given] A simple topology question

    Hi all again.

    I have a very simple question that I cannot find a clue.
    \mathrm{Ext}(A):=(\overline{A})^{c} (external of A), where \overline{\cdot} is the closure and \cdot^{c} is the complementary.
    \partial A:=\overline{A}\backslash\overset{\circ}{A} (boundary of A).
    Prove that \partial\mathrm{Ext}(A)=\partial A.

    Thanks.
    Last edited by bkarpuz; March 11th 2009 at 12:40 PM.
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  2. #2
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    A point x \in \partial (A) if and only if each open set that contains x contains a point in A and a point in A^c.
    Read that definition carefully several times.
    It follows from that definition that any such point is a boundary point of both A and A^c, recall that \left[ {A^c } \right]^c  = A.
    If X is the whole space then X = \text{Int} (A) \cup \partial (A) \cup \text{Ext} (A).
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  3. #3
    Senior Member bkarpuz's Avatar
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    Exclamation

    Thanks Plato, but I guess I need more help.

    For instance, do we have \mathrm{Int}(\overline{A})=\mathrm{Int}(A)?
    I could not find a counter example.

    If so, I have a solution as follows.
    \partial(\mathrm{Ext}(A))=\partial((\overline{A})^  {c})
    .............._ =\overline{(\overline{A})^{c}}\backslash\mathrm{In  t}((\overline{A})^{c})
    .............._ =(\mathrm{Int}(\overline{A}))^{c}\backslash(\overl  ine{(\overline{A})})^{c}
    .............._ =(\mathrm{Int}(\overline{A}))^{c}\backslash(\overl  ine{A})^{c}
    .............._ \stackrel{?}{=}(\mathrm{Int}(A))^{c}\backslash(\ov  erline{A})^{c}
    .............._ =\overline{A^{c}}\backslash\mathrm{Int}(A^{c})
    .............._ =\partial(A^{c})
    .............._ =\partial(A).

    Is that true?

    Note. We work on \mathbb{R}.
    Last edited by bkarpuz; March 3rd 2009 at 09:31 AM. Reason: Note added.
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  4. #4
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    A slight variation on what Plato said: if you define the interior points of a set, A, as you do in metric topology- points, p, such that some neighborhood of p is a subset of A- then you can define the exterior of A as the set of all interior points of the complement of A and the boundary points of A as the set of all points in the space that are neither interior points nor exterior points of A. Because "complement of complement of A" is A, The interior points of A are the exterior points of A^c and vice versa. What does that tell you about the boundary points of A.
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  5. #5
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    I don't know why in the world you are making it so complicated.
    In fact, I do not follow some of those steps.
    \begin{gathered}<br />
  A = (0,1) \cup (1,2)\; \Rightarrow \;\text{Int} (A) = A \hfill \\<br />
  \overline A  = \left[ {0,2} \right]\; \Rightarrow \;\text{Int} \left( {\overline A } \right) = (0,2) \ne \text{Int} (A) \hfill \\ <br />
\end{gathered}
    Last edited by Plato; March 3rd 2009 at 11:25 AM. Reason: correction
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  6. #6
    Senior Member bkarpuz's Avatar
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    Talking

    May be this is just because of my field (applied mathematics), I always wish to see some computations instead of words.
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  7. #7
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    Quote Originally Posted by Plato View Post
    I don't know why in the world you are making it so complicated.
    In fact, I do not follow some of those steps.
    \begin{gathered}<br />
  A = (0,1) \cup (1,2)\; \Rightarrow \;\text{Int} (A) = A \hfill \\<br />
  \overline A  = \left[ {0,2} \right]\; \Rightarrow \;\text{Int} \left( {\overline A } \right) = (0,2) \ne \text{Int} (A) \hfill \\ <br />
\end{gathered}
    So to continue that example, \partial A = \{0,1,2\}, \text{Ext}(A) = (-\infty,0)\cup(2,\infty) and \partial(\text{Ext}(A)) = \{0,2\} \ne \partial A.

    In fact, as Plato showed, a set and its complement always have the same boundary. But a set and its "external" need not have the same boundary.
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  8. #8
    Senior Member bkarpuz's Avatar
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    Exclamation

    Thank you both Plato and Opalg, I really appreciate this help.
    I could never find that counter example.
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  9. #9
    Senior Member bkarpuz's Avatar
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    Cool [Solved: A counter example is given]

    Okay, I have a counter example too.
    I cannot belive how I could not see it earlier.
    Let A:=[0,1]\cap\mathbb{Q}, then we have \partial A=[0,1] but \partial\mathrm{Ext}A=\{0,1\}.
    This example also escapes from the gap that (0,1)=\mathrm{Int}\overline{A}\neq\mathrm{Int}A=\e  mptyset.

    Again many thanks for your asistance.
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