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.
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8$\begingroup$ 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. $\endgroup$– Qiaochu YuanCommented Sep 7, 2012 at 5:28
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$\begingroup$ Sorry, I edit to include definitions. $\endgroup$– Pedro PerezCommented Sep 7, 2012 at 6:15
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1$\begingroup$ Take the finite complement topology on any infinite set. $\endgroup$– Evan JenkinsCommented Sep 7, 2012 at 6:39
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4$\begingroup$ That space is not US. $\endgroup$– Pedro PerezCommented Sep 7, 2012 at 7:53
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3$\begingroup$ By the way, what do KC and US stand for? I imagine KC means "kompact closed", but US is puzzling me. ("unique sequence"?) $\endgroup$– Henry CohnCommented Sep 7, 2012 at 12:32
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3 Answers
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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).
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$\begingroup$ pual Fabel, your nbohd around z is not clear $\endgroup$– 00GBCommented Jan 17, 2022 at 3:40
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1$\begingroup$ Let U be an arbitrary open dense subspace of [0,1], with the relative topology. For example U=[0,1/2) union (1/2,1). The set z union U is a typical open set in X which contains z. $\endgroup$ Commented Jan 18, 2022 at 18:29
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1$\begingroup$ Pual Fabel, Is $X\cup \{z\}$ still compact? $\endgroup$– 00GBCommented Jan 19, 2022 at 4:09
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$\begingroup$ Yes, given arbitrary open cover, one of the open sets U contains z. The complement of U is closed in the compact space [0,1], and hence can be covered by finitely many of the surviving open sets in the original cover. $\endgroup$ Commented Jan 19, 2022 at 4:40
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1$\begingroup$ Paul Fabel, I was asking since I found the question, see math.stackexchange.com/q/4358229/707884, the user asked question has to do with your question and in the comments, they mentioned to your answer but I did know not how easy can one modify your example if it is possible $\endgroup$– 00GBCommented Jan 19, 2022 at 4:59
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Start with the one point compactification of the minimal uncountable well ordered space and then split the maximum point into two points.
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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 $K \subset X$.
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1$\begingroup$ +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. $\endgroup$ Commented Sep 7, 2012 at 14:02