Let and . Show that is an open cover of .
So choose . Pick such that . This implies that . Is this correct? Could I pick in a different way?
What if we had a finite subset ?
The statement that set is compact means every open covering has a finite subcover.
Having one open cover which has a finite subcover does not imply that the set is compact.
That is clearly an open covering of and the collection contains a finite subcover.
But is still not compact.