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Math Help - Help! Advanced Calculus

  1. #1
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    Angry Help! Advanced Calculus

    Let K be a sequentially compact subset of R^n and suppose O is an open set of R^n that contains K. Prove that there is some positive number r such that for any point U in K, Ball (U) with radius of r is contained in O.

    Hint: by contradiction.
    Thank you!
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  2. #2
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    Here is an outline.

    Suppose NOT: n \in Z^ +  \quad  \Rightarrow \quad \left( {\exists k_n } \right)\left[ {B\left( {k_n ;1/n} \right) \not\subset O} \right].
    Because K is sequentially compact <br />
\left( {k_n } \right) has an accumulation point (limit point) is K. Call it k.
    Because O covers K, \left( {\exists \varepsilon  > 0} \right)\left[ {B\left( {k;\varepsilon } \right) \subset O} \right].
    \left( {\exists N \in Z^ +  } \right)\left[ {\frac{1}{N} < \frac{\varepsilon }{4}} \right]\quad \& \quad \left( {\exists J \in Z^ +  ,J > N} \right)\left[ {k_J  \in B\left( {k;\varepsilon/2 } \right)} \right].
    But \left( {\exists t \in K} \right)\left[ {t \in B\left( {k_J ;1/J} \right)\backslash O} \right]
    d(t,k) \le d\left( {t,k_J } \right) + d\left( {k_j ,k} \right) < \frac{\varepsilon }{4} + \frac{\varepsilon }{2} < \varepsilon .

    Do you see the contradiction there?
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  3. #3
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    Thank you very much!
    It is still difficult for me, but I will follow your idea.
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  4. #4
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    Quote Originally Posted by violetsf View Post
    It is still difficult for me, but I will follow your idea.
    May I suggest that you look into some textbooks for the following topics: \varepsilon \mbox{-nets}, totally bounded, and/or precomapct. Each of these is close to what the given problem is asking.
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  5. #5
    Super Member Rebesques's Avatar
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    Good job P.


    Now here's another way to see this, using the properties of distance functions.

    Define the distance function d(y,\partial O)=\inf_{x\in \partial O}|x-y|, \ y\in K,  \ \partial O denoting the boundary of O (which is nonempty as O is open). This is continuous as the norm is continuous. If we show that \inf_{y\in K}d(y,\partial O)=d_0>0, then for all radii r<d_0 and all y\in K we will have B(y,r)\subset O.

    Suppose on the contrary \inf_{y\in K}d(y,\partial O)=0. Then there is a sequence y_n\in K such that \lim d(y_n,\partial O)=0. Since K is sequentially compact, the sequence (y_n) has a subsequence (y_{k_n}) that converges to some element y_0\in K.

    So we get 0=\lim d(y_{k_n},\partial O)=d(y_0,\partial O). This means y_0\in \partial O, as \partial O is closed; This contradicts the fact that y_0\in K\subset O.
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