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Thread: Totally bounded set

  1. #1
    Nov 2008

    Totally bounded set

    I try to prove that every subset of a totally bounded set is totally bounded, but I am not really sure with my argument here. Could someone help me check it?

    Suppose A is a subset of a totally bounded set X. Since X is totally bounded, there exists an \frac{\epsilon}{2}-net F=\{x_1,x_2,..,x_n\} that covers X, i.e X\subset \bigcup \{B(x_i,\epsilon/2):x_i\in F\}. Since A is a subset of X, A is also a subset of the union above. So, A\cap B(x_i,\epsilon/2) for some x_i \in F. Let y\in A\cap B(x_i,\epsilon/2). Then B(y,\epsilon) covers A since this ball contains every ball with radius \epsilon/2 above. Something does not sound right here since I am not sure if the new ball with radius \epsilon contains all the balls with radius \epsilon/2
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
    Nov 2010
    Madrid, Spain
    Removing all balls which do not contain points of A and rearranging the subindices if necessary: A\subset\bigcup_{i=1}^{m}B(x_i,\epsilon/2)\;(m\leq n) .

    For all i=1,\ldots ,m there exists one point z_i\in A\cap B(x_i,\epsilon/2). Then, B(x_i,\epsilon/2)\subset B(z_i,\epsilon) .

    In fact, if y\in B(x_i,\epsilon/2) we have:

    d(y,z_i)\leq d(y,x_i)+d(x_i,z_i)<\epsilon/2+\epsilon/2=\epsilon

    So, A\subset \bigcup_{i=1}^{m}B(x_i,\epsilon/2)\subset \bigcup_{i=1}^{m}B(z_i,\epsilon) hence, A is totally bounded.
    Last edited by FernandoRevilla; Mar 10th 2011 at 12:53 AM.
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