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

A topological space is called a $KC$-space provided that every compact subset is closed.

So Hausdorff spaces imply $KC$-spaces and $KC$-spaces imply $US$-spaces.

I would like to know:

Is there an example that shows $US$-space does not imply $KC$-space?

Why is a hereditarily Lindelöf $KC$ - space $(X,\tau)$ Katetov - $KC$ if and only if there is a weaker $US$-topology $\sigma \subset \tau?

Why is $X=[0,\omega_1]$ not sequential?

  • $\begingroup$ See mathoverflow.net/questions/106571/… for the first question. $\endgroup$ – dfeuer Jul 27 '13 at 19:04
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    $\begingroup$ It should be obvious that Kansas City spaces are US spaces and that New York City spaces are US spaces that are not KC spaces. $\endgroup$ – Ross Millikan Jul 27 '13 at 21:48

A space $X$ is sequential it every sequentially closed set in $X$ is closed, where $A\subseteq X$ is sequentially closed iff $A$ contains the limit of every sequence in $A$ that converges in $X$. $X=[0,\omega_1]$ is not sequential because the subset $A=[0,\omega_1)$ is sequentially closed but not closed. It’s clearly not closed, since $\omega_1$ is a limit point of $A$. To see that $A$ is sequentially closed, let $\langle\xi_n:n\in\omega\rangle$ be a sequence in $A$ converging to some $\xi\in X$, and let $\eta=\sup_{n\in\omega}\xi_n$. Then $\eta<\omega_1$, so $(\eta,\omega_1]$ is an open nbhd of $\omega_1$ disjoint from $\{\xi_n:n\in\omega\}$, so $\omega_1\ne\xi$, and therefore $\xi\in A$.

Now let $p$ be any point not in $X$, and let $Y=X\cup\{p\}$. For each $\alpha<\omega_1$ let $(\alpha,\omega_1)\cup\{p\}$ be a basic open nbhd of $p$. In other words, $[0,\omega_1)\cup\{p\}$ is homeomorphic to $X$: to form $Y$ from $X$ I’ve simply split $\omega_1$ into two points, $\omega_1$ and $p$. You can easily check that $Y$ is compact, assuming that you know why $X$ is compact. $X$ is a compact subset of $Y$ that is not closed in $Y$, so $Y$ is not $KC$. Finally, if a sequence $\langle y_n:n\in\omega\rangle$ in $Y$ converges to $\omega_1$, it is eventually constant at $\omega_1$ (meaning that there is an $m\in\omega$ such that $y_n=\omega_1$ for all $n\ge m$), and if it converges to $p$, it is eventually constant at $p$. This implies that a sequence that converges to $\omega_1$ or to $p$ does not converge to any other point of $Y$. Finally, $Y\setminus\{\omega_1,p\}$ is Hausdorff, and sequences that converge to points of $Y\setminus\{\omega_1,p\}$ are eventually in $Y\setminus\{\omega_1,p\}$, so $Y$ is $US$.

Added: It is not true that a hereditarily Lindelöf $KC$ space is Katětov $KC$ iff it admits a weaker $US$ topology, even if weaker is understood to mean strictly weaker. Example $3.4$ of Chiara Baldovino & Camillo Constantini, ‘On some questions about $KC$ and related spaces’ is a countable $KC$ space $\langle X,\sigma\rangle$ that is not Katětov $KC$. Plainly $\langle X,\sigma\rangle$ is hereditarily Lindelöf, and it’s not hard to show that $\langle X,\sigma\rangle$ has a strictly weaker $US$ topology.

Specifically, in the notation of the paper fix any $\varphi:\Bbb N\to\Bbb N$, and replace the original fundamental system of open nbhds of $p_2$ by the smaller system $\{V_{2,\varphi,n}:n\in\Bbb N\}$. If $\sigma'$ is the resulting topology, the only sequences in $X$ that are convergent in $\sigma'$ but not in $\sigma$ are sequences that converge to $p_2$ in $\sigma'$, and they converge only to $p_2$ in $\sigma'$. Thus, $\langle X,\sigma'\rangle$ is $US$.


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