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May 13, 2019 at 14:18 history edited Zhenchao Lyu CC BY-SA 4.0
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Apr 25, 2019 at 15:43 comment added Zhenchao Lyu Another natural question: Is every countable complete lattice equipped Scott topology sober?
Mar 8, 2019 at 19:10 comment added Zhenchao Lyu 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.
Sep 25, 2018 at 15:12 history asked Zhenchao Lyu CC BY-SA 4.0