Thread
:
Show a closed subset of a compact set is also compact
View Single Post
Dick
#
5
Oct29-09, 12:46 PM
Sci Advisor
HW Helper
Thanks
P: 25,235
Since F is compact, there is a finite subcover. So that subcover also covers E. Now you don't need the S\E set to cover E. What's left is a finite subcover of the original cover that covers E.