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A Heyting algebra is a distributive bounded lattice equipped with a special operation "$\to$" obeying $(c\land a)\leq b \iff c\leq (a\to b)$. These algebras are of interest since they serve as models of intuitionist logic. A complete Heyting algebra is a Heyting algebra which is complete as a lattice. My question is whether every Heyting algebra can be embedded into a complete Heyting algebra, while preserving finite joins, meets, and the "$\to$" operation.

This seems like a basic question about Heyting algebras, but I couldn't find any sources. One would think the completion of Heyting algebras would be common knowledge, were it true. Maybe there's an obvious (dis)proof I just haven't thought of?

I've already tried the obvious: the Dedekind-Macneille completion lets us embed any Heyting algebra into a complete, bounded, distributive lattice. In the completion, we can define $(a\to b) := \bigvee\{c : (c\land a)\leq b\}$, and I was able to prove that the "$\to$" operation is respected by the embedding. Unfortunately, I was not able to show that the completion is a Heyting algebra, as I couldn't prove the axiom $c\leq (a\to b) \iff (c\land a)\leq b$ holds in the completion.

I tried a slightly different completion, but had a similar problem where I couldn't prove it's a Heyting algebra. I think there's a way to get a completed Heyting algebra using a larger completion than Dedekind-Macneille, but it would be nice to know whether this is already a solved problem.

Jade Vanadium
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    When you say preserving joins and meets, do you mean preserving finite joins and finite meets (which always exist in a Heyting algebra) or preserving all joins and meets that already happen to exist? – S.C. Jan 08 '25 at 18:38
  • @S.C. EDIT: only the finite joins and finite meets need to be preserved. In the standard Heyting algebra $H$, given by propositional formulae modulo intuitionistic equivalence, the join of infinitely many propositional variables is $\top$, which is intuitively wrong. The Dedekind-Macneille completion preserves that join, which is bad since it results in a violation of the Heyting algebra axioms. Infinitary meets also shouldn't be preserved, since in $H$ the meet of all variables is $\bot$; even the Dedekind-Macneille completion fails to preserve that meet (which is good). – Jade Vanadium Jan 08 '25 at 19:38
  • The canonical extension of a Heyting algebra is again a Heyting algebra. It doesn't preserve any "inherently infinite joins or meets", which seems to serve you. Here, an inherently infinite join (respectively meet) is one which no finite subset gets the same value. It preserves all finite ones. I'm not sure I can't find a source soon enough, but the canonical extension can be obtained as the family of order-ideals of the Priestley space of the original algebra. (Or perhaps the Esakia space?) – amrsa Jan 08 '25 at 20:05
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    The MacNeille completion of a Heyting algebra is known to again be a Heyting algebra; see this paper: https://citeseerx.ist.psu.edu/document?repid=rep1&type=pdf&doi=3abd9064b45c5f9bed828bfb6ac4e05b970ec485. The canonical extension is also a completion, as discussed in papers by Gehrke and Bezhanishvili. Occasionally there are other completions that can be used in special cases, e.g., for a countable Heyting algebra, one can take a refinement of the canonical extension which preserves a designated countable family of joins, which can be broadly called the "Rasiowa-Sikorski" extension. – Rodrigo Nicolau Almeida Feb 24 '25 at 08:45
  • @RodrigoNicolauAlmeida You're right, I just checked my notes where I was working with a particular completion I came up with, and I made a mistake when identifying it with Dedekind-Macneille. I've removed that error from my question. – Jade Vanadium Feb 24 '25 at 13:39

1 Answers1

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I found a reference.

In [1], page 145, there's the paragraph:

Following [2] (here 2 refers to this small bibliography), the canonical extension of a Heyting algebra $H$ is a complete Heyting algebra $H^{\sigma}$ such that there is a Heyting embedding $e:H\to H^{\sigma}$ for which, as a lattice, $H^{\sigma}$ is the canonical extension of $H$. Canonical extensions of Boolean algebras are defined similarly.

The paper is about characterizing those Heyting algebras for which the canonical extension and the profinite completion coincide (this is the case for Boolean algebras and for bounded distributive lattices).

From the first paragraph of the introduction:

It is possibly folklore that the profinite completions and canonical extensions of bounded distributive lattices and of Boolean algebras coincide. However, they can be different for Heyting algebras, and our aim in this paper is to give a criterion to determine when they coincide. A description of the canonical extension $H^{\sigma}$ of a Heyting algebra $H$ in terms of the Priestley dual ($=$ Esakia dual) of $H$ is known. We provide a similar description of the profinite completion $\hat{H}$ of $H$, and we use these descriptions to obtain our criterion. In addition, we give a simple algebraic proof that the profinite completion and canonical extension of a bounded distributive lattice and of a Boolean algebra coincide.

They proceed to define the profinite completion of Heyting algebras. Thus you have at least two such completions.

[1] G. Bezhanishvili, M. Gehrke, R. Mines, P. Morandi, Profinite completions and canonical extensions of Heyting algebras, Order 23 (2006), p 143-161.

[2] M. Gehrke, B. Jónsson, Monotone bounded distributive lattices expansions, Math. Jpn. 52 (2000), p 197-213.

amrsa
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  • This is great, thanks! It'll take me a few days to get to reading these, but do you happen to know whether either of those completions are minimal in any sense? – Jade Vanadium Jan 09 '25 at 15:48
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    @JadeVanadium Generally, the minimal completion of a poset or lattice is de Dedekin-MacNeille completion. If any of these is minimal with respect to being a completion of a Heyting algebra (in the sense that the original is embeddable into these) I don't know. Actually I didn't check in detail the reference [1]; perhaps they say if one is embeddable into the other when they do not coincide. By the way, I don't have access to reference [2]. The canonical extension is seldom a minimal completion because of the compactness condition. – amrsa Jan 09 '25 at 16:24