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A fundamental theorem by Stone asserts that any Boolean algebra is isomorphic to a subalgebra of the archetypical Boolean algebras, that is the power sets of a set $X$ (equipped with intersection, union and complementation).

I was wondering whether a similar result carries over to Heyting algebras, that is whether it is true or not that any (complete) Heyting algebras is isomorphic to a subalgebra of the Heyting algebra given by the open subsets of a topological space. If this is not the case (as I suspect), is there any prototype of Heyting algebra which every Heyting algebra (complete or not) can be proven to be isomorphic to?

Thank you in advance.

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  • $\begingroup$ @MarcoVergura: Just a remark: If every Boolean algebra was isomorphic to some powerset, then there would be no countable Boolean algebras. $\endgroup$
    – Kyle Gannon
    Jul 14, 2014 at 17:33
  • $\begingroup$ The difficulty, in some sense, is the Heyting implication. Every Heyting algebra is a distributive lattice, and every distributive lattice can be embedded in a complete Heyting algebra, but this is an embedding of distributive lattices, not Heyting algebras. On the other hand, not every distributive lattice can be embedded as a subalgebra of the distributive lattice of open subsets of a topological space – this is the question of whether there are "enough points". $\endgroup$
    – Zhen Lin
    Jul 14, 2014 at 17:57
  • $\begingroup$ Have you checked Dunn and Hardegree's Algebraic Methods in Philosophical Logic? On p. 385, they claim that every Heyting algebra is isomorphic to a Heyting algebra of open sets. Is this what you're looking for? $\endgroup$
    – Nagase
    Jul 25, 2014 at 5:22
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    $\begingroup$ Have you checked in Helena Rasiowa & Roman Sikorski, The Mathematics of Metamathematics (1963), Ch.IV. Pseudo-Boolean algebras, page 128 : §3. Representation theorems. "The following theorem explain the connection between pseudo-Boolean algebras and topological Boolean algebras: For every pseudo-boolean algebra $A$ there exists a topological Boolean algebra $B$ such that $A = \mathfrak S(B)$ [McKinsey and Tarski, 1946]" ? $\endgroup$
    – Mauro ALLEGRANZA
    Aug 29, 2014 at 14:24

1 Answer 1

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Esakia Duality might just be the thing that you are looking for.

For every Heyting algebra $A$ there exists a so called Esakia space $\mathscr{X}=(X, \leq, \mathscr{O})$, which is a certain kind of ordered topological space, such that $A$ is isomorphic to the Heyting algebra of clopen up-sets of $\mathscr{X}$.

This gives a dual equivalence between the category of Heyting algebras and the category of Esakia spaces, very similar to the well-know Stone duality between Boolean algebras and Stone spaces. However the category of Esakia spaces is not a full subcategory of the category of ordered topological spaces and continuous and order preserving functions between them.

Finally you can give a purely topological description of the category of Esakia spaces, in the sense that it is isomorphic to a (non full) subcategory of the category of Spectral spaces and spectral maps see Bezhanishvili et al. 2010 Theorem 7.12.

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