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Math Help - A finite subset of a metric space

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    A finite subset of a metric space

    Show that a finite subset of a metric space has no limit points and is therefore a closed set.
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    Quote Originally Posted by aliceinwonderland View Post
    Show that a finite subset of a metric space has no limit points and is therefore a closed set.
    A subset of a metric space is closed if and only if it contains its limit points.
    To be a limit point any neighborhood needs to contains infinitely many points within the set.
    Thus, a finite subset cannot have any limit points.
    Thus, the set of limit points is the empty set.
    But, the finite subset contains the empty set.
    Thus, the finite subset is closed.
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    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by aliceinwonderland View Post
    Show that a finite subset of a metric space has no limit points and is therefore a closed set.
    This is an alternate solution to TPHs...it is a little more formal. Note, you may want to wait for confirmation of my solution by a more knowledgable member


    Let X be a metric space and let \left\{p\right\}=A\subset X. Now the only possibilities for a limit point of A is the point p, but note that since there is only one point A no matter how small you make an open ball around p there will not be another point of A, thus a singleton in X has no limit points, thus it is closed and consequently its complement is open. So now let another set \left\{x_1,x_2,\cdots,x_n\right\}=B\subset X where n is finite. It is clear that B=\bigcup_{i=1}^{n} x_i we will now use the following lemma

    Lemma: The intersection of any finite number of open sets is itself open.

    Proof: Let U_i~~1\leqslant i\leqslant n be open, then let x\in\bigcap_{i=1}^{n}U_i. This implies that x\in U_i~\forall i, and since each U_i is open there exists an open ball \mathcal{O}_{\varepsilon_i}(p,X) corresponding to each sU_i such that \mathcal{O}_{\varepsilon_i}(p,X)\subset U_i. So now let \varepsilon=\min\left\{\varepsilon_1,\varepsilon_2  ,\cdots\right\}, then \mathcal{O}_{\varepsilon}(x,X)\subset\bigcap_{i=1}  ^{n}U_i\quad\blacksquare


    Using the above lemma and the fact that the complement of any singleton in X is open we can see that the complement of any finite point set \left[\bigcup_{i=1}^{n}x_i\right]^c=\bigcap_{i=1}^{n}x_i^c is open, thus the finite set is closed.
    Last edited by Mathstud28; January 13th 2009 at 12:02 PM. Reason: Detail
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