# Uniqueness of limits and compactness implies closure

It is not difficult to prove that in a Hausdorff topological space every compact set is closed, and almost trivial that if in a topological space X every compact set is closed then X is T1. As uniqueness of limits for a topological space lies between the T1 and T2 properties, it is a candidate for a necessary and sufficient condition for every compact set to be closed. Is it necessary or sufficient?

• When you say "uniqueness of limits", do you mean the statement that if a sequence converges, then it converges uniquely? Alternatively, please explain the precise statement you are referring to. – user44191 Dec 13 '18 at 17:10
• Yes, when I say uniqueness of limits I mean the property of a topological space that every convergent sequence converges uniquely. – Daniel Elessar Dec 13 '18 at 17:11
• It's not going to be sufficient, because sequence convergence isn't enough to detect "limits". You should be able to get a counterexample from a "line with two origins" construction on the uncountable ordinal $\omega_1 + 1$. – Nate Eldredge Dec 13 '18 at 18:01
• I think it's necessary, though, because a convergent sequence together with one of its limits is a compact set, and it is closed only if it contains all the limits of the sequence. – Nate Eldredge Dec 13 '18 at 18:03
• @user44191: That suggests looking at a Tychonoff plank with two corners; apparently the corner is a limit point, but not the limit of any ordinal-indexed sequence. On the other hand, if you say "ultrafilter limits are unique" or "net limits" then this is equivalent to being Hausdorff. – Nate Eldredge Dec 13 '18 at 18:19

A proof can be found here, but I'll recap the idea in case the link goes away: suppose $$x_n \to x$$ and $$x_n \to y$$, we need to show that $$x=y$$.
Suppose $$x \neq y$$. One of the sets $$N(x):=\{n : x_n = x\}$$ or $$N(y):= \{n: x_n = y\}$$ is not cofinite, say the second is not. Omit all terms $$x_n$$ with $$x_n =y$$ and so create a new sequence $$y_n$$ that still converges to $$x$$ and $$y$$. The set $$C:=\{y_n\} \cup \{x\}$$ is compact (hence closed as we assume $$X$$ is KC) but $$y$$ is also in the closure of $$C$$ but not in $$C$$, contradiction and thus $$x=y$$, and $$X$$ is US.