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Thread: Compact Sets.

  1. #1
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    Compact Sets.

    Let $\displaystyle (X,d_{x})$ be a continuous function. Suppose that $\displaystyle K\subset X$ is compact and that $\displaystyle A\subset\ \mathbb{R}^k$ is compact.
    Prove that $\displaystyle f(K)\bigcap A\ =\ \emptyset$ implies dist$\displaystyle (f(K),A)>0$

    It's analysis and there isn't an extra class for that unlike the other modules (even though it's considered the hardest).
    Thnx in advance
    Last edited by mr fantastic; Dec 10th 2011 at 08:08 PM. Reason: Title.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: Compact Sets.

    Quote Originally Posted by maxgunn555 View Post
    Let $\displaystyle (X,d_{x})$ be a continuous function. Suppose that $\displaystyle K\subset X$ is compact and that $\displaystyle A\subset\ \mathbb{R}^k$ is compact.
    Prove that $\displaystyle f(K)\bigcap A\ =\ \emptyset$ implies dist$\displaystyle (f(K),A)>0$

    It's analysis and there isn't an extra class for that unlike the other modules (even though it's considered the hardest).
    Thnx in advance

    You know that $\displaystyle f(K)$ is compact, and so $\displaystyle f(K)\times A$ is compact, so the mapping $\displaystyle d:f(K)\times A\to\mathbb{R}$ obtains a minimum value $\displaystyle d(f(k_0),a_0)$ for some $\displaystyle (f(k_0),a_0)\in f(K)\times A$. Clearly though $\displaystyle d(f(k_0),a_0)>0$ otherwise $\displaystyle f(k_0)=a_0$ which contradicts $\displaystyle f(K)\cap A=\varnothing$.
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