Edited sentence: Consider  Observe that B is a directed set under inclusion and for each  for all  Consider the net  I claim that this net cannot have a convergent subnet since for each x there exists  but for all  we have that  and thus is not in some open neighborhood of x.Conversely, suppose that every net in X has a convergent subnet.First suppose that X is compact.For every  define.For the sake of contradiction, let  be an open cover of X with no finite subcover.Hence, the net  meets U infinitely often and so has a subnet that converges to x.Let A be a directed set and  be a net in X.Let I be any set and  be a collection of closed subsets of X such that  for each finite.Let x be an element of this intersection.Otherwise,  would be an open cover for X with no finite subcover.The collection  has the property that every finite subcollection has non-empty intersection.Then  as well.Then, each open neighborhood U of x meets  for each.This is a contradiction and completes the proof.Thus, by the remark above, we have that.Thus, for each   there exists   with.We will need the following observation (see  Finite intersection property)