A COMPACT SPACE IS NOT ALWAYS SI-COMPACT

Abstract

We consider the poset $\mathcal{𝒬}(P)$ of all nonempty compact saturated subsets of the Scott space of a poset $P$, equipped with the reverse inclusion order. We also introduce the notion of $T$-lattices, that is, a complete lattice $L$ is called a $T$-lattice if for any $x\in L\setminus \{1_L\}$, $\uparrow x\setminus \{x\}\in \mathcal{𝒬}(L)$. We prove that if $P$ and $Q$ are quasicontinuous domains or $T$-lattices, $P$ is order isomorphic to $Q$ if and only if $(\mathcal{𝒬}(P),\supseteq )$ is order isomorphic to $(\mathcal{𝒬}(Q),\supseteq )$. We also prove that for a compact space $X$, if the Scott space of the open set lattice $\mathcal{𝒪}(X)$ is sober, then $X$ is $SI$-compact. Using this result and the Isbell example of a non-sober complete lattice, we present a compact sober space which is not $SI$-compact. This gives a positive answer to an open problem posed by Zhao and Ho.