Alexandrov topology and Scott topology
20 Mar
2003
20 Mar
'03
2:26 p.m.
Hello, A common example of a Heyting Algebra is one defined on an arbitrary topology. In the cases of an Alexandrov topology and a Scott topology, are either of these Boolean Algebras (special case of a HA) or purely Heyting Algebras? Regards, Bill Halchin
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Galchin Vasili