I'm working on this problem, and I'm not sure how to solve it.
Let be a topological space and let be countably compact. Prove that every subset of that is closed relative to is also countably compact.
I have a similar proof where this problem is stated without the 'countably' part. I'm unsure how to proceed with countably compact sets... Any help would be appreciated.
Ah, so it is similar to the proof with compactness... I'll just change the proof a little, then. If you could verify that it is correct I would grateful.
Suppose that is countably compact and let be closed. We will let be a countable cover of . Since is open, it follows that along with is a countable open cover of . We know that is countably compact, so can be covered by a finite number of sets from , , and possibly . Since , covers . Therefore, is countably compact.