How can I show this set is compact?

Question

Let X be a topological space and let x_n be a sequence in X that converges to some x. Show that the set {x, x1, x2,...x_n,....} is compact. The problem is it's not assumed X is a metric space.
Thank you

Answer 1

Since x_n --> x, for every neighborhood V of x there exists k such that n >= k => x_n is in V. This is valid in any topological space, not only metric or metrizable ones.

From this definition, it follows that x_n --> x if, and only if, every neighborhood of x contains all but a finite number of terms of (x_n).

Let {V} be any open cover of A = {x, x1, x2,...x_n,....}. Then, x belongs to some V of {V}. Being open, V is a neighborhood of x. Since x_n --> x, V contains all but a finite number of terms x_n. Therefore, V contains all but a finite number (possibly 0) of elements of A. Since the elements of A that aren't in V form, therefore, a finite set, this set can be covered by a finite subcollection {V1, V2...Vm} of {V}. Therefore, {V, V1,V2....V_m} is a finite subcollection of {V} that covers A.

Since this holds for every open cover of A, it follows A is compact.

Answer 2

The key is the fact that the sequence converges. I almost said "the set you describe contains all its limit points" but that merely makes it a closed set.

I will have to get my topology book out when I get home.

If you can cover X with only finitely many subsets of X then you have compactness, but I would still have to refer to my topology book for the details.