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Math Help - [SOLVED] Sequentially Compact

  1. #1
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    [SOLVED] Sequentially Compact

    Problem:
    Let K be a sequentially compact subset of \mathbb{R}^n and suppose that O(this is the letter 'O') is an open subset of \mathbb{R}^n that contains K. Prove that there exists some positive number r such that for any point  u \in K ,  B_{r} (u) \subseteq O

    Proof:
    (I have a hard time with these "prove that there exists..." type proofs)

    Let O be is an open subset of \mathbb{R}^n that contains K, which is sequentially compact.
    Pick a sequence  x_{n} \in \mathbb{R}^n such that  \forall n \in N and  u \in K such that  u \in B_{\frac{1}{n}}(x_{n}) since \left( \frac{1}{n} > 0 \right)

    Since K is sequentially compact, then there is a subsequence  x_{n_k} \mapsto u .
    Let \frac{1}{n} < \frac{r}{2}

    then

     B_{\frac{1}{n}} \left( x_{n_k} \right) \subseteq B_{r}(u) \subseteq O
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  2. #2
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    Quote Originally Posted by Paperwings View Post
    Problem:
    Let K be a sequentially compact subset of \mathbb{R}^n and suppose that O(this is the letter 'O') is an open subset of \mathbb{R}^n that contains K. Prove that there exists some positive number r such that for any point  u \in K ,  B_{r} (u) \subseteq O
    Assume this is false. Then for r = \frac{1}{n} we can choose \bold{u}_n \in \mathcal{K} such that B_{\frac{1}{n}} (\bold{u}_n)\not \subseteq \mathcal{O}. Therfore, \{\bold{u}_n\} is a sequence in \mathcal{K} - a sequencially kompact set. Thus, we can find a convergent subsequence \bold{x}_n \in \mathcal{K} such that there is r_n>0 (with r_n \to 0) so that B_{r_n}(\bold{x}_n) \not \subseteq \mathcal{O}. We shall show this leads to a contradiction. Let \bold{x} = \lim \bold{x}_n. Since \mathcal{K} is closed it follows \bold{x} \in \mathcal{K}\subset \mathcal{O}. Now since \mathcal{O} is open there is \epsilon > 0 such that B_{\epsilon}(\bold{x}) \subseteq \mathcal{O}. Since \bold{x}_n \to \bold{x} it means there is N such that if |\bold{x}_n - \bold{x}| < \epsilon for n\geq N. Now \bold{x}_n \to \bold{x} and r_n eventually gets smaller than \epsilon. Thus there is a K so that B_{r_K}(\bold{x}_K)\subseteq B_{\epsilon}(\bold{x}) \subseteq \mathcal{O}.

    This is a contradiction.
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  3. #3
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    Thank you TPH.
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