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Math Help - Prove the inverse function of a compact set is compact.

  1. #1
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    Prove the inverse function of a compact set is compact.

    Problem: Let f:R^n -> R^m be continuous. Let K be a compact subset of R^m. Suppose there exists r > 0 such that ||f(u)|| >= r||u|| for each u in R^n. Suppose that f^{-1} (K) is not an empty set, prove it is compact.

    My proof:

    Now f^{-1}(K) = {x in R^n : f(x) in K}, with f^{-1}(K) maps f(K) to R^n.

    Let {Uk} be a sequence in f^{-1}(K), then there exists r > 0 such that ||f(Uk)|| >= r||Uk||.

    Now, f(Uk) is in K, K is compact implies K is bounded, implies there exists a number M such that M >= ||f(Uk)|| >= r||Uk|| > ||Uk||, thus shows {Uk} is bounded, implies f^{-1}(K) is bounded.

    Furthermore, suppose the sequence {Yk} in f^{-1}(K) converges to the point Y, then for each {Yk} in R^n, we have f(Yk) in K. Now f(Yk) -> f(Y) because f is continuous, and f(Y) is in K because K is closed, furthermore, f(Y) is in the set of f^{-1}(K).

    Thus proves f^{-1}(K) is closed, and we conclude that it is compact since it is closed and bounded.

    Q.E.D.

    PLEASE CHECK, THANK YOU!!!

    KK
    Last edited by tttcomrader; April 6th 2007 at 06:49 PM.
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  2. #2
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    Please review the statement of the problem you posted. It seems to me there are some internal inconsistencies there. In what you posted as a proof, you wrote “||Uk|| > Uk”. That has no possible meaning. Here is why: ||Uk|| is a number, but Uk is a point on a metric space R^m.

    I will say that I have not seen any similar problem. That makes me suspicious of the exact statement.
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  3. #3
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    Actually, there is no reason for me to even have ||Uk|| > Uk, because once I have M>=||Uk||, I have already proven Uk is bounded, so it is deleted now.
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