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Math Help - Totally bounded subset

  1. #1
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    Totally bounded subset

    I need to prove the following:

    Suppose M is a metric space and A \subseteq M. Then A is totally bounded if and only if, for every \epsilon >0, there is a finite \epsilon-dense subset of A.

    (\Rightarrow ): Let \epsilon > 0. Then  \epsilon > 0 . So there exist sets A_1, A_2, \cdots, A_k, with diam(A_i)<\epsilon and  A=\displaystyle \cup_{i=1}^{k} A_i. Now \forall i let a_i \in A_i. Let X=\{ a_i \} _{i=1}^{k}. So X is a finite subset of A. I claim that X is \epsilon-dense in A. To see this, let a \in A. I must show that \exists some element of X so that \rho (a,x)< \epsilon. If a \in A, then a \in A_i for some i. Since a_i \in A_i, diam(A_i) < \epsilon, \rho (a, a_i) < \epsilon.

    (\Leftarrow ): Now suppose that for \epsilon > 0, A has a finite \epsilon-dense subset. I must prove that A is totally bounded. Since A has a finite \epsilon-dense subset, \exists \ a \in A so that \rho (x,a)< \epsilon. So diam(A)= l.u.b. \{ \rho (a_1, a_2)|a_1, a_2 \in A \}< \epsilon so diam(A)< \epsilon \ \forall a.

    Now for the problem: I am not sure if I have included all the definition for totally bounded; that is, I think I still have another step, but am not sure. Thanks for your help.
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  2. #2
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    You cannot conclude that \text{diam}(A)\le \epsilon. Because that is not true.

    Can you show that A\subseteq \bigcup\limits_{k = 1}^n {\mathcal{B}\left( {a_k ;\epsilon } \right)}~?
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  3. #3
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    Quote Originally Posted by Plato View Post
    You cannot conclude that \text{diam}(A)\le \epsilon. Because that is not true.

    Can you show that A\subseteq \bigcup\limits_{k = 1}^n {\mathcal{B}\left( {a_k ;\epsilon } \right)}~?
    Since A is \epsilon-dense in M, \ \forall a \in A \ \exists \ y \in B so that \rho (a,y)< \epsilon, which means A\subseteq \bigcup\limits_{k = 1}^n {\mathcal{B}\left( {a_k ;\epsilon } \right)} .
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  4. #4
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    The statement that X is totally bounded means that there exist an \epsilon \text{-net} for X.

    If you have a finite \epsilon \text{-dense} set on X then is that an \epsilon\text{-net}?
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  5. #5
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    We have not used the term \epsilon-net in my class. I am in an advanced calculus class. I know that if X is totally bounded, then X can be covered by a finite number of subsets of M whose diameters are all less than \epsilon. Is that what you mean by \epsilon-net?
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  6. #6
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    Since there is an \epsilon-dense set in A, say  \{ x_1, x_2, \cdots, x_n \} is \epsilon-dense in A, then B[x_i; \epsilon], \cdots, B[x_n; \epsilon] form a covering of A by sets of diameter < \epsilon. Hence A is totally bounded. QED.

    Right?
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  7. #7
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    Well that is what I would say is correct.
    I have never seen the term \epsilon \text{-dense}.
    I was guessing that it meant \epsilon\text{-net} which is finite.
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