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Math Help - Closure of a totaly bounded set is totally bounded

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    Closure of a totaly bounded set is totally bounded

    I think I have the idea in my head, but I don't know if I am expressing it clearly in math terms.

    The closure of S, \bar{S} is defined to be S \cup S' (where S' are the limit points of S)

    S totally bounded implies that \forall \epsilon > 0 \; \exists finite number of points x_1, ... , x_n \in S s.t

    S = \cup_{k=1}^n \; B(x_k,\epsilon)

    So we already have all the points in S, i.e the union of the balls is a cover for S. We just need to wory about S', the limit points.

    x \in S' \rightarrow \forall \epsilon > 0 \;, \exists y \in S s.t d(x,y) < \epsilon

    so if any limit point is less than \epsilon away from a point in S. We can say

    \bar{S} = \cup_{k=1}^n \; B(x_k,2\epsilon)

    Right? Is my thinking correct? The limit points are close to the set so we can make sure we get them by making the radius bigger.
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    Super Member girdav's Avatar
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    I think it's ok.
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    Quote Originally Posted by Jame View Post
    The closure of S, \bar{S} is defined to be S \cup S' (where S' are the limit points of S)
    S totally bounded implies that \forall \epsilon > 0 \; \exists finite number of points x_1, ... , x_n \in S s.t
    \bar{S} = \cup_{k=1}^n \; B(x_k,2\epsilon)
    Some instructors can be very picky here.
    You want to show that for each \varepsilon >0 there is an \varepsilon\text{-net} for \overline{S}.

    That means if we find an \frac{\varepsilon}{2}\text{-net} for S and apply your idea it works.
    Last edited by Plato; April 8th 2011 at 01:38 PM.
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    Oh I see.

    So we start with

    S = \cup_{k=1}^n \; B(x_k,\epsilon/2)

    and apply the same argument to get


    \bar{S} = \cup_{k=1}^n \; B(x_k,\epsilon)
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    Quote Originally Posted by Jame View Post
    Oh I see.
    So we start with
    S = \cup_{k=1}^n \; B(x_k,\epsilon/2)
    and apply the same argument to get
    \bar{S} = \cup_{k=1}^n \; B(x_k,\epsilon)
    Exactly. But it needs be \bar{S} \subseteq \cup_{k=1}^n \; B(x_k,\epsilon)
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    My book defines totally bounded with an equal sign. Honestly "contained" makes much more sense though, it seems to be defined that way in every other source I've looked about. Is there a reason for this discrepancy in my book?


    EDIT:

    I think the whole space X is equal to the union, while a subset of X is contained.

    Thanks for all your help!
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    Quote Originally Posted by Jame View Post
    My book defines totally bounded with an equal sign. Honestly "contained" makes much more sense though, it seems to be defined that way in every other source I've looked about.
    If S were the whole space S=\overline{S} and there is nothing to prove.
    I know of no textbook that uses that definition.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Jame View Post
    My book defines totally bounded with an equal sign. Honestly "contained" makes much more sense though, it seems to be defined that way in every other source I've looked about. Is there a reason for this discrepancy in my book?


    EDIT:

    I think the whole space X is equal to the union, while a subset of X is contained.

    Thanks for all your help!
    Exactly. It is much more common in my experience that the definition involves equality instead of containment (despite what you indicate). This is just the difference of point of view where one usually thinks of topological (and in this case metric) spaces as living in ambient spaces (or not). So, if you're discussing the topology of a fixed space ( \mathbb{R} say) then since all the topology is done within this space a set X contained in your ambient space can be said to be covered by \left\{U_\alpha\right\}_{\alpha\in\mathcal{A}} if \displaystyle X\subseteq\bigcup_{\alpha\in\mathcal{A}}U_\alpha. But, if you think of X as a space in it's own right (with the subspace topology say) then saying \displaystyle X\subseteq\bigcup_{\alpha\in\mathcal{A}}U_\alpha amounts to equality since (remembering now that we're thinking of X as the ambient space itself) we have U_\alpha\subseteq X for every \alpha\in\mathcal{A}.
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