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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?

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    $\begingroup$ 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. $\endgroup$ – user44191 Dec 13 '18 at 17:10
  • $\begingroup$ Yes, when I say uniqueness of limits I mean the property of a topological space that every convergent sequence converges uniquely. $\endgroup$ – Daniel Elessar Dec 13 '18 at 17:11
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    $\begingroup$ 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 $\omega_1 + 1$. $\endgroup$ – Nate Eldredge Dec 13 '18 at 18:01
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    $\begingroup$ 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. $\endgroup$ – Nate Eldredge Dec 13 '18 at 18:03
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    $\begingroup$ @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. $\endgroup$ – Nate Eldredge Dec 13 '18 at 18:19
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A KC space is indeed US. This is classical. I believe its already in the paper where these notions (KC = all compact subsets are closed, US = all convergent sequences have unique limits) are defined (this paper by Wilansky, I believe).

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.

This old question and its answers give ideas to find a US space that is not KC.

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