Compact metric space proof?

How do I show that any closed subset K of a compact metric space M is also compact?

I want to make use of the definitions of compactness and covering.

Thank you in advance!

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  • 8 years ago
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    Take any open cover of K, and put in one additional open set: the complement of K.

    You now have an open cover of M. This gives you a finite subcover, which if it includes the complement of K that can be eliminated, and you now have a finite open cover of K which is a subcover of the original open cover of K.

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