University of Birmingham > Talks@bham > Theoretical computer science seminar > When exactly is Scott sober?

When exactly is Scott sober?

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  • UserWeng Kin Ho, National Institute of Education, Singapore
  • ClockFriday 18 June 2010, 14:00-15:00
  • HouseUG40 Computer Science.

If you have a question about this talk, please contact Paul Levy.

To every T_0 space X, one can induce a partial order called its specialization order. If, in addition, X is sober, then this specialization order on X turns out to be a dcpo, i.e., a poset in which every directed subset has a supremum. Moreover, the topology on X is contained in the Scott topology on the specialization order of X. Because of this, it has been asked if every dcpo is always sober under its Scott topology. To this question, a negative answer was given by P.T. Johnstone in a humorously-titled 1981 paper ’’Scott is not always sober’’. Thus, a natural problem is to find an order-characterization of those dcpo’s whose Scott topology is sober. This talk addresses this problem.

This talk is part of the Theoretical computer science seminar series.

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