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 May 24, 2023, 1:34 pm ad1c9bdddfhttps://brainmass.com/math/real-analysis/real-analysis-27182
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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.
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