# A space in which sequences have unique limits but compact sets need not be closed

A topological space is KC if every compact subspace is closed. A topological space is US if every convergent sequences has exactly one limit. Does someone know an easy example of a US space which is not KC? Thanks.

• These terms happen to both 1) be terrible search terms and 2) have unguessable meanings if you're not familiar with them, so you might want to include definitions. – Qiaochu Yuan Sep 7 '12 at 5:28
• Sorry, I edit to include definitions. – Pedro Perez Sep 7 '12 at 6:15
• Take the finite complement topology on any infinite set. – Evan Jenkins Sep 7 '12 at 6:39
• That space is not US. – Pedro Perez Sep 7 '12 at 7:53
• By the way, what do KC and US stand for? I imagine KC means "kompact closed", but US is puzzling me. ("unique sequence"?) – Henry Cohn Sep 7 '12 at 12:32

To create a counterexample X, start with the closed interval [0,1] (with the usual topology) and attach a new point z whose neighborhoods are open dense subsets of [0,1].

Observe [0,1] is a compact nonclosed subspace of X and thus X is not a KC space. However no sequence in [0,1] converges to z and in particular all convergent sequences in X have unique limits.

The finite complement topology on an infinite set does not yield a counterexample since every infinite sequence converges to every point of the space.

In general no counterexample Y can be a sequential space since if Y is a sequential space then Y is a KC space iff Y is a US space. ( Recall Y is a sequential space if every nonclosed set B contains a convergent sequence whose limit lies outside B).

Start with the one point compactification of the minimal uncountable well ordered space and then split the maximum point into two points.

In COROLLARY 1 of it, $X^+$ denotes the one point compactification of the topological space $X$:

COROLLARY: Let $X$ be a Hausdorff space.Then:

(a) $X^+$ is always $US$.

(b) $X^+$ is $KC$ if and only if $X$ is a $\kappa$ space.

So it suffices to choose a Hausdorff space $X$, which is not $\kappa$ space. then $X^+$ is $US$ but is not $KC$.

PS: The topological space $X$ is called $\kappa$ space, if A subspace $A$ is closed in $X$ if and only if $A \cap K$ is closed in $K$, for all compact subset $K \subset X$.

• +1 for the link. Theorem 1 says: $T_2 \implies KC \implies US \implies T_1$ and none of the implications reverses even for compact spaces. – Ramiro de la Vega Sep 7 '12 at 14:02