Logic-Compactness in complete lattice

2025-12-17T03:13:16Z

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 $\mathcal{C}$ satisfies 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.)

Answer by Emil Jeřábek for Logic-Compactness in complete lattice

2025-12-17T09:34:55Z

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.

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.

If $\bigwedge\Gamma$ exists for every nonempty $\Gamma\subseteq L$, and $L$ is logic-compact, then $L$ is well founded.

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:

$\{\bigwedge\Gamma_0:\varnothing\ne\Gamma_0\subseteq\Gamma\text{ finite}\}\subseteq L$ 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.
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.
By fiat.

Logic-Compactness in complete lattice - MathOverflow

Logic-Compactness in complete lattice

Answer by Emil Jeřábek for Logic-Compactness in complete lattice