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!
- euleriaLv 68 years agoFavorite Answer
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.