12
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ @MarcoVergura: Just a remark: If every Boolean algebra was isomorphic to some powerset, then there would be no countable Boolean algebras. $\endgroup$ Jul 14 '14 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 '14 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 '14 at 5:22
  • 1
    $\begingroup$ 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]" ? $\endgroup$ Aug 29 '14 at 14:24
10
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.