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Basic notions: $1$, A partially ordered set is a dcpo if each of its directed subsets has a supremum. (https://en.wikipedia.org/wiki/Complete_partial_order)\

$2$, A subset O of a dcpo P is called Scott-open if it is an upper set and if it is inaccessible by directed joins, i.e. if all directed sets D with supremum in O have non-empty intersection with O. The Scott-open subsets of a dcpo P form a topology on P, the Scott topology.(https://en.wikipedia.org/wiki/Scott_continuity)\

When is the Scott topology on a dcpo sober? This is a very basic and interesting question in Domain Theory. Johnstone firstly gave an example of dcpo whose Scott topology is not sober(https://link.springer.com/book/10.1007%2FBFb0089899#page=294).

Recently, I have read a short paper "COMPLETION OF A CONSTRUCTION OF JOHNSTONE" by Isbell (http://www.ams.org/journals/proc/1982-085-03/S0002-9939-1982-0656096-4/S0002-9939-1982-0656096-4.pdf). In the paper he gave a complete lattice whose Scott topology is not sober. But the example is hard to understand, so I am wondering if there is an easier example for this.

By the way, there is an open problem: Find a distributive lattice whose Scott topology is not sober.(Solved)

(https://en.wikipedia.org/wiki/Domain_theory. For more details, refer to"Domain Theory" http://www.cs.bham.ac.uk/~axj/pub/papers/handy1.pdf and the red book"Continuous Lattices and Domains"https://www.amazon.com/Continuous-Lattices-Encyclopedia-Mathematics-Applications/dp/0521803381).

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    $\begingroup$ The problem about finding a distributive lattice whose Scott topology is not sober has been solved in a recent preprint "A complete Heyting algebra whose Scott space is non-sober". arxiv.org/abs/1903.00615 But essentially they use Isbell's example. $\endgroup$ Commented Mar 8, 2019 at 19:10
  • $\begingroup$ Another natural question: Is every countable complete lattice equipped Scott topology sober? $\endgroup$ Commented Apr 25, 2019 at 15:43

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