US does not imply AB We say that a topological space $X$ is:


*

*$AB$, provided that $X$ is $T_1$ and for each pair $(A, B)$ of compact, disjoint subsets of $X$ there is $U$, an open subset of $X$, such that either $A \subseteq U$ and $U \cap B = \varnothing$ or $B \subseteq U$ and $U \cap A = \varnothing$.

*$US$, provided that convergent sequences have a unique limit.


I would like to know an example of a topological space that is $US$ but not $AB$.
Thanks
 A: Let $S=\{2^{-n}:n\in\Bbb N\}$, and let $K=\{0\}\cup S$. Let $X=\{0,1\}\times K$. Points of $\{0,1\}\times S$ are isolated. For $i\in\{0,1\}$ let $p_i=\langle i,0\rangle$, and let $\mathscr{U}$ be a free ultrafilter on $\Bbb N$. A set $V\subseteq X$ is a nbhd of $p_i$ iff 


*

*$p_i\in V$,  

*$(\{i\}\times S)\setminus V$ is finite, and  

*$\{n\in\Bbb N:\langle 1-i,2^{-n}\rangle\in V\}\in\mathscr{U}$.


For $i\in\{0,1\}$ let $K_i=\{i\}\times K$; $K_i$ inherits the same topology from $X$ as it does from $\Bbb R^2$, so it’s compact. However, $\operatorname{cl}K_i=K_i\cup\{p_{1-i}\}$, so neither of $K_0$ and $K_1$ has a nbhd disjoint from the other. Thus, $X$ is not an $AB$-space.
To see that $X$ is a $US$-space, note first that $p_0$ and $p_1$ are the only non-isolated points of $X$ and hence the only possible limit points of non-trivial sequences. Now suppose that $A\subseteq S_i$ for some $i\in\{0,1\}$. If $A$ is finite, $p_{1-i}$ has a nbhd disjoint from $A$. If $A$ is infinite, let $$N_A=\{n\in\Bbb N:\langle i,2^{-n}\rangle\in A\}\;.$$ Since $\mathscr{U}$ is a free ultrafilter, there is a $U\in\mathscr{U}$ such that $N_A\setminus U$ is infinite. Let
$$V=K_{1-i}\cup\{\langle i,2^{-n}\rangle:n\in U\}\;;$$
then $V$ is an open nbhd of $p_{1-i}$ such that $A\setminus V$ is infinite. It follows that no sequence with infinitely many terms in $S_i$ can converge to $p_{1-i}$. From this it follows easily that no sequence in $X$ converges to both $p_0$ and $p_1$ and hence that $X$ is a $US$-space.
