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    Real analysis

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    Show that if K is compact and F is closed then K intersection F is compact.

    © BrainMass Inc. brainmass.com March 4, 2021, 6:04 pm ad1c9bdddf
    https://brainmass.com/math/real-analysis/real-analysis-27182

    Solution Preview

    Proof:

    Since K is compact, then any sequence of K can have a convergent subsequence which converges to an element in K. Now we ...

    Solution Summary

    This is a proof regarding compact and closed sets.

    $2.49

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