A countable metric space whose Hoare power space is not co-sober

Xu asked whether the Hoare power space of a co-sober space, and in particular of a $T_2$-space or a metric space, must be co-sober. We give a negative answer. He and Zhao recently constructed a countable $T_1$ space $X$ whose lattice of open sets has a non-sober Scott topology. We observe that their topology has a countable subbase. The classical Ponomarev construction then yields an open continuous surjection from a countable zero-dimensional metrizable space $M$ onto $X$. Direct and inverse images then make the Scott space $\mathrm{\Sigma }\mathcal{O}(X)$ a retract of $\mathrm{\Sigma }\mathcal{O}(M)$, so $\mathrm{\Sigma }\mathcal{O}(M)$ is not sober. Finally, we show that, for every $T_0$-space $Y$, the Hoare power space $\text{PH}(Y)$ is co-sober if and only if the Scott space $\mathrm{\Sigma }\mathcal{O}(Y)$ is sober. Consequently, $\text{PH}(M)$ is not co-sober. Therefore, the same example answers negatively a question of He and Zhao about Scott sobriety of the lattices of open sets of countable Hausdorff spaces.