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Thread: [SOLVED] Sequentially Compact

  1. #1
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    [SOLVED] Sequentially Compact

    Problem:
    Let K be a sequentially compact subset of $\displaystyle \mathbb{R}^n$ and suppose that O(this is the letter 'O') is an open subset of $\displaystyle \mathbb{R}^n$ that contains K. Prove that there exists some positive number r such that for any point $\displaystyle u \in K $ , $\displaystyle 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 $\displaystyle \mathbb{R}^n$ that contains K, which is sequentially compact.
    Pick a sequence $\displaystyle x_{n} \in \mathbb{R}^n $ such that $\displaystyle \forall n \in N $ and $\displaystyle u \in K $ such that $\displaystyle u \in B_{\frac{1}{n}}(x_{n}) $ since $\displaystyle \left( \frac{1}{n} > 0 \right)$

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

    then

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

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