A space in which sequences have unique limits but compact sets need not be closed - MathOverflow most recent 30 from http://mathoverflow.net2013-05-26T07:33:21Zhttp://mathoverflow.net/feeds/question/106571http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/106571/a-space-in-which-sequences-have-unique-limits-but-compact-sets-need-not-be-closedA space in which sequences have unique limits but compact sets need not be closedPedro Perez2012-09-07T05:16:48Z2012-09-09T04:56:49Z
<p>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.</p>
http://mathoverflow.net/questions/106571/a-space-in-which-sequences-have-unique-limits-but-compact-sets-need-not-be-closed/106583#106583Answer by Paul Fabel for A space in which sequences have unique limits but compact sets need not be closedPaul Fabel2012-09-07T08:25:56Z2012-09-07T08:25:56Z<p>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].</p>
<p>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.</p>
<p>The finite complement topology on an infinite set does not yield a counterexample since every infinite sequence converges to every point of the space.</p>
<p>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).</p>
http://mathoverflow.net/questions/106571/a-space-in-which-sequences-have-unique-limits-but-compact-sets-need-not-be-closed/106584#106584Answer by Paul Fabel for A space in which sequences have unique limits but compact sets need not be closedPaul Fabel2012-09-07T09:00:46Z2012-09-07T09:00:46Z<p>Start with the one point compactification of the minimal uncountable well ordered space and then split the maximum point into two points.</p>
http://mathoverflow.net/questions/106571/a-space-in-which-sequences-have-unique-limits-but-compact-sets-need-not-be-closed/106597#106597Answer by AliReza Olfati for A space in which sequences have unique limits but compact sets need not be closedAliReza Olfati2012-09-07T11:28:10Z2012-09-09T04:56:49Z<p>I refer to <strong>COROLLARY 1</strong> of <a href="http://www.jstor.org/discover/10.2307/2316017?uid=3738280&uid=2&uid=4&sid=21101169814551" rel="nofollow">This Article.</a></p>
<p>In <strong>COROLLARY 1</strong> of it, $X^+$ denotes the one point compactification of the topological space $X$:</p>
<blockquote>
<p><strong>COROLLARY</strong>: Let $X$ be a Hausdorff space.Then:</p>
<p>(a) $X^+$ is always $US$.</p>
<p>(b) $X^+$ is $KC$ if and only if $X$ is a $\kappa$ space.</p>
</blockquote>
<p>So it suffices to choose a Hausdorff space $X$, which is not $\kappa$ space. then $X^+$ is $US$ but is not $KC$.</p>
<p>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 $K \subset X$.</p>