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Assume $\Gamma$ is an infinite set composed of formulas (of finite length), and $A$ is a formula (of finite length). For example, $\Gamma=\{x,y\sqcup z,x_1,x_2,x_3,\cdots\}$, $A=(x\sqcap y)\sqcup(x\sqcap z)$. Obviously, $\mathop⨅\Gamma\le A$ holds in any of the complete distributive lattices $\langle L,\le\rangle$ (i.e. $\mathop⨅\Gamma\le A$ holds for any $x,y,z,x_1,x_2,x_3,\cdots\in L$). Meanwhile, we also can find a finite subset $\{x,y\sqcup z\}\subseteq\Gamma$ such that $\mathop⨅\{x,y\sqcup z\}\le A$ holds in any of the complete distributive lattices $\langle L,\le\rangle$ (i.e.$\mathop⨅\{x,y\sqcup z\}\le A$ holds for any $x,y,z\in L$).

Let $\mathcal{C}$ be a kind of complete lattice (with extra operations). $\mathcal{C}$ is logic-compact iff $\mathcal{C}$ satisfy the following property:

For any infinite set $\Gamma$ composed of formulas (of finite length), for any formula $A$ (of finite length), if $\mathop⨅\Gamma\le A$ holds in any instance $\mathcal{L}$ of $\mathcal{C}$, then we can always find a finite subset $\Gamma_0\subseteq\Gamma$ such that $\mathop⨅\Gamma_0\le A$ holds in any instance $\mathcal{L}$ of $\mathcal{C}$. (The operators in the formulas mentioned above contain both those in the lattice and extra equipped.)

Another example: the infinite inequality $x\sqcap(x\to y)\sqcap x_1\sqcap x_2\sqcap x_3\cdots\le y$ holds in any of the Heyting algebras, which can be attributed to the finite inequality $x\sqcap(x\to y)\le y$ holding in any of the Heyting algebras. I'm wondering whether any infinite inequality holding in any of the algebra can be attributed to a finite inequality holding in any of the algebra.

So, my question is: Does any kind of complete lattice with any operation have logic-compactness? If not, what conditions do the complete lattice and the operations need to satisfy for having the property? (e.g., bounded, distributive, etc.)

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    $\begingroup$ If the lattice is not complete, how is the definition of logic compactness involving $\bigsqcap\Gamma$ intended to be interpreted in the first place? $\endgroup$ Commented yesterday
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    $\begingroup$ Simultaneously cross-posted at math.stackexchange.com/questions/5115256/… . Please, do not do that. $\endgroup$ Commented 22 hours ago
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    $\begingroup$ You are making the question more and more unclear. I assume “kind of lattice” just means a class $C$ of lattices, in which case the only way I can read “instance” of “this kind” is that is refers to lattices $L\in C$. But then saying “$\bigwedge\Gamma\le y$ holds for all $L\in C$” makes no sense whatsoever, as $\Gamma$, $\bigwedge\Gamma$, and $y$ only belong to one particular lattice. And you still did not clarify what happens when the meet of $\Gamma$ does not exist. $\endgroup$ Commented 18 hours ago
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    $\begingroup$ Hmm, so the question actually asks something completely different from what it looked originally. However, it is still fairly unclear. First, “formulas” are things like $\forall x\,\exists y\,x<y$ that are either true or false; they do not evaluate in any meaningful way to elements of the lattice. It seems that when you write “formulas”, you actually mean “terms over the lattice signature”, which do indeed take values in the lattice. Certainly $y\lor z$ and $(x\land y)\lor(x\land z)$ are terms. (I’m not going to use the square symbols as they are difficult to type.) However, “$x\to y$” ... $\endgroup$ Commented 13 hours ago
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    $\begingroup$ ... is not a term over lattices, as there is no $\to$ operation in the language of lattices. So I’m not sure what to make of that. Second, despite having been prompted for it already several times, you still haven’t explained what is supposed to happen WHEN $\bigwedge\Gamma$ DOES NOT EXIST in the lattice in question. This is absolutely crucial. For example, you claim that $\bigwedge\{x,y\lor z,x_1,x_2,x_3,\dots\}\le(x\land y)\lor(x\land z)$ holds in distributive lattices. But $(\mathbb Q,\le)$ is a distributive lattice, and when I take for $x_1,x_2,\dots$ a descending rational sequence ... $\endgroup$ Commented 13 hours ago

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While the question is still unclear, it seems now that it is very different from the question answered below, which was based on my best guess from reading the first revision.

Since MathJax does not seem to support \bigsqcap anyway, I’ll write meets using the standard $\bigwedge$ notation.

Logic compactness is essentially equivalent to well foundedness:

Proposition. Let $L$ be a lattice.

  1. If $L$ is well founded, then for every nonempty $\Gamma\subseteq L$, $\bigwedge\Gamma$ exists, and equals $\bigwedge\Gamma_0$ for some finite $\Gamma_0\subseteq\Gamma$. Thus, $L$ is logic-compact.
  2. If $\bigwedge\Gamma$ exists for every nonempty $\Gamma\subseteq L$, and $L$ is logic-compact, then $L$ is well founded.
  3. If $\bigwedge\Gamma$ does not necessarily exist, then I do not know how to interpret the definition of logic compactness in the first place.

Proof:

  1. $\{\bigwedge\Gamma_0:\varnothing\ne\Gamma_0\subseteq\Gamma\text{ finite}\}\subseteq L$ is nonempty, hence it has a minimal element $\bigwedge\Gamma_0$ by well foundedness. Then $\bigwedge\Gamma_0$ is $\bigwedge\Gamma$, i.e., $\bigwedge\Gamma_0\le x$ for each $x\in\Gamma$, lest $\bigwedge(\Gamma_0\cup\{x\})<\bigwedge\Gamma_0$, contradicting minimality.

  2. If $L$ is not well founded, there is a strictly decreasing chain $x_0>x_1>x_2>\cdots$. Put $\Gamma=\{x_n:n\in\omega\}$ and $y=\bigwedge\Gamma$. Then $y<\bigwedge\Gamma_0=\min\Gamma_0\in\Gamma$ for any nonempty finite $\Gamma_0\subseteq\Gamma$, witnessing that $L$ is not logic-compact.

  3. By fiat.

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    $\begingroup$ Thank you for your kind and conscientious reply! Please forgive my ambiguous expressions; there was some ambiguity in my question, indeed. I have modified some of my expressions again, which should be crucial. $\endgroup$ Commented 19 hours ago
  • $\begingroup$ If I understand logic compactness, it is exactly the lattice property that every element is dually compact. This is equivalent to well-foundedness if the lattice is complete, but need not be equivalent otherwise. $\endgroup$ Commented 3 hours ago
  • $\begingroup$ @KeithKearnes As transpired in the later revisions, the property is intended to be something completely different (to begin with, it is a property of classes rather than individual lattices). I think I’ll just delete this answer. $\endgroup$ Commented 29 mins ago

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