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Thread: Compact sets

  1. #1
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    Compact sets

    Quote Originally Posted by Plato View Post
    I assume that we are in a metric space.
    Here are traditional proofs for both properties.
    Suppose that x is a limit point of K but x\notin K.
    \left( {\forall y \in K} \right) this is true r_y  = \frac{{d(x,y)}}{4} > 0.
    The collection \left\{ {B(y;r_y )} \right\}_{y \in K} covers K.
    So finite subcollection K \subset \bigcup\limits_{j = 1}^n {B(y_j ;r_{y_j } )} also covers K.

    But note that x \in \bigcap\limits_{j = 1}^n {B(x ;r_{y_j } )}. That is a open set that contains x and no other point of K. Contradiction.


    For bounded, there is finite collection \bigcup\limits_{j = 1}^n {B(y_j ;1)} covering K.
    Let M = \max \left\{ {d(y_k ,y_j )} \right\} + 2. It is easy to show that M is a bound for K.
    I don't understand the bolded part
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  2. #2
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    Re: Compact sets

    Hey Akatsuki.

    Open sets don't contain the limit points of the set. Does that help?
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  3. #3
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    Re: Compact sets

    Quote Originally Posted by Akatsuki View Post
    I don't understand the bolded part
    This is the standard proof for compactness.
    Cover each point with a ball and cover $x$ with a corresponding ball so the two ball do not intersect.
    Now compactness gives a finite collection of balls.
    Union of open sets is open and finite intersection of balls is open.
    Thus we produce and open set that contains $x$ and on other point of $K$.
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