Abstract Monads prove to be useful mathematical tools in theoretical computer science, notably in denoting different effects of programming languages. In this paper, we investigate a type of monads which arise naturally from Keimel and Lawson’s $\mathbf{K}$ -ification. A subcategory of $\mathbf{TOP}_{\mathbf{0}}$ is called of type $\mathrm{K}^{*}$ if it consists of monotone convergence spaces and is of type $\mathrm K$ in the sense of Keimel and Lawson. Each such category induces a canonical monad $\mathcal K$ on the category $\mathbf{DCPO}$ of dcpos and Scott-continuous maps, which is called the order- $\mathbf{K}$ -ification monad in this paper. First, for each category of type $\mathrm{K}^{*}$ , we characterize the algebras of the corresponding monad $\mathcal K$ as k-complete posets and algebraic homomorphisms as k-continuous maps, from which we obtain that the order- $\mathbf{K}$ -ification monad gives the free k-complete poset construction over the category $\mathbf{POS}_{\mathbf{d}}$ of posets and Scott-continuous maps. In addition, we show that all k-complete posets and Scott-continuous maps form a Cartesian closed category. Moreover, we consider the strongness of the order-K-ification monad and conclude with the fact that each order-K-ification monad is always commutative.

Abstract The study of the sobriety of Scott spaces has got a relatively long history in domain theory. Lawson and Hoffmann independently proved that the Scott space of every continuous directed complete poset (usually called domain) is sober. Johnstone constructed the first directed complete poset whose Scott space is non-sober. Soon after, Isbell gave a complete lattice with a non-sober Scott space. Based on Isbell’s example, Xu, Xi, and Zhao showed that there is even a complete Heyting algebra whose Scott space is non-sober. Achim Jung then asked whether every countable complete lattice has a sober Scott space. The main aim of this paper is to answer Jung’s problem by constructing a countable complete lattice whose Scott space is non-sober. This lattice is then modified to obtain a countable distributive complete lattice with a non-sober Scott space. In addition, we prove that the topology of the product space $\Sigma P\times \Sigma Q$ coincides with the Scott topology of the product poset $P\times Q$ if the set Id(P) and Id(Q) of all incremental ideals of posets P and Q are both countable. Based on this, it is deduced that a directed complete poset P has a sober Scott space, if Id(P) is countable and $\Sigma P$ is coherent and well filtered. In particular, every complete lattice L with Id(L) countable has a sober Scott space.