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

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

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

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

    So, $\displaystyle A\subset \bigcup_{i=1}^{m}B(x_i,\epsilon/2)\subset \bigcup_{i=1}^{m}B(z_i,\epsilon)$ hence, $\displaystyle A$ is totally bounded.
    Last edited by FernandoRevilla; Mar 9th 2011 at 11:53 PM.
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