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Thread: closure

  1. #1
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    closure

    Show that $\displaystyle X- E^{\circ} = \overline{X-E} $ where X is a metric space and E is a subset of X. Can we do this without using DMorgan's Laws (e.g. without using intersection of closed sets)?

    Because you are taking away $\displaystyle E^{\circ} \subset E $ from $\displaystyle X $. Whereas $\displaystyle \overline{X-E} = X-E \cup (X-E)' $ where $\displaystyle (X-E)' $ is the set of limit points of $\displaystyle X-E $. So you are taking away more points from $\displaystyle X $. Adding back the limit points gives you $\displaystyle X- E^{\circ} $ (e.g. you are not adding any interior points).
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  2. #2
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    Gdansk, Poland
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    First of all we should notice two facts:

    1.
    $\displaystyle X \setminus E^\circ \supset X \setminus E$

    2.
    $\displaystyle X \setminus \overline{X \setminus E} \subset (X \setminus X \setminus E) = E$

    Ad 1.
    It follows from $\displaystyle E^\circ \subset E$

    Ad 2.
    It follows from $\displaystyle X \setminus E \subset \overline{X \setminus E}$

    Now, since $\displaystyle E^\circ$ is open, $\displaystyle X \setminus E^\circ$ is closed.
    Since closure of A is the smallest closed set containing A from 1 we have:
    $\displaystyle X \setminus E^\circ \supset \overline{X \setminus E}$

    Now, since $\displaystyle \overline{X \setminus E}$ is closed, $\displaystyle X \setminus \overline{X \setminus E}$ is open.
    Since interior of A is the largest open set in A we from 2 we have

    $\displaystyle X \setminus \overline{X \setminus E} \subset E^\circ$

    Thus: $\displaystyle \overline{X \setminus E} \supset X \setminus E^\circ$

    So we have: $\displaystyle \overline{X \setminus E} = X \setminus E^\circ$
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