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A topological space is called a $US$-space provided that each convergent sequence has a unique limit.

We know that for locally compact spaces,‎ ‎$ ‎T‎_{2} ‎‎‎\equiv KC‎‎$.

We have: ‎‎$ ‎T_2‎ ‎\Rightarrow ‎KC‎‎‎ ‎\Rightarrow ‎US‎\Rightarrow ‎T_1‎ $‎.

Is it possible to say "for locally compact spaces, $ ‎US‎\Rightarrow T_2$? ( It means $T_2 \equiv US$)

Martin Sleziak
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fatemeh
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  • Not unless $US \Rightarrow KC$ whatever $KC$ means – Henrique Tyrrell Aug 14 '13 at 17:01
  • So, to make a counterexample, try $[0,\omega_1)$ with two limit points added at the top... – GEdgar Aug 14 '13 at 19:27
  • Please, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. – Martin Sleziak Aug 15 '13 at 08:19

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GEdgar has given one example in the comments. Start with the ordinal space $\omega_1$, and add two points, $p$ and $q$. For each $\alpha<\omega_1$ let $U_\alpha(p)=\{p\}\cup(\alpha,\omega_1)$ and $U_\alpha(q)=\{q\}\cup(\alpha,\omega_1)$, and take $\{U_\alpha(p):\alpha<\omega_1\}$ and $\{U_\alpha(q):\alpha<\omega_1\}$ as local bases at $p$ and $q$, respectively. The resulting space $X$ is not $T_2$, since $p$ and $q$ do not have disjoint nbhds, but it is compact, locally compact by any definition, and $US$. ($X$ is $US$ because $\omega_1$ is $T_2$, and the only sequences converging to $p$ or to $q$ are trivial ones.)

The same idea can be applied to $\beta\omega$. Fix $p\in\beta\omega\setminus\omega$, let $q$ be a new point not in $\beta\omega$, and let $X=\beta\omega\cup\{q\}$. Topologize $X$ by making $\beta\omega$ an open subset of $X$ with its usual topology and making $U\subseteq X$ an open nbhd of $q$ iff $q\in U$, and $\{p\}\cup\big(U\setminus\{q\}\big)$ is an open nbhd of $p$ in $\beta\omega$. (In other words, we make $q$ a second copy of $p$.) The only convergent sequences in $X$ are the trivial ones, so $X$ is $US$, and it’s clear that $X$ has the other required properties.

Brian M. Scott
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