# 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

## 3 Answers

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.

-

I refer to COROLLARY 1 of This Article.

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