# What do you call a lattice whose meet operation preserves disjointness of subsets?

To make my question more precise and compact (and probably more intuitive), let me define the following:

A subset $S$ of a lattice is mutually disjoint if for each $x \in S$, $\bigvee(S - \lbrace x \rbrace)$ is defined and $x \wedge \bigvee(S - \lbrace x \rbrace) = \varnothing$.

If for every two mutually disjoint subsets $S_1$ and $S_2$ of $L$, $S_1 \wedge S_2 = \lbrace s_1\wedge s_2 \mid s_1 \in S_1, s_2 \in S_2\rbrace$ is also mutually disjoint, we say that $\wedge$ of $L$ preserves disjointness.

Now my question is: What do you call a lattice with this property? Is this property equivalent to a well-known property?

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I'm having trouble understanding the question. Is $S - x$ supposed to be $S - \{x\}$, i.e., the complement of $\{x\}$ relative to $S$? Are we assuming the lattice has arbitrary joins? –  Todd Trimble Oct 28 '10 at 20:55
@Todd: the question is completely clear. $S$ is finite, $S-x=S\setminus\{x\}$. @Tunococ: Unfortunately I do not know the answer. I have never seen this before. I suggest that if you really need an answer, send a message to Ralph McKenzie (Vanderbilt) or to J.B. Nation (Hawaii). –  Mark Sapir Oct 28 '10 at 22:31
Mark, the question wasn't clear; he either forgot to say the lattice (not $S$, mind you) was finite or that he meant to say sup-lattice. (And, if it wasn't clear to me, then surely you don't mind if I seek clarification? Why do you address me in such a supercilious manner?) –  Todd Trimble Oct 29 '10 at 1:06
@Todd: I do not know the word "supercilious". But it probably means something negative. Nothing negative was intended and I am sorry if you read it that way. I still think that the question was clear enough, but certainly you have the right to ask for clarifications. By the way, why is finiteness of $L$ needed? I thought only $S$ must be finite. In any case, the question is not answered. –  Mark Sapir Oct 29 '10 at 1:31
Okay, thanks Mark. And you're right, maybe he means $S$ is finite. In fact that seems likely now. –  Todd Trimble Oct 29 '10 at 1:43

Many lattices do have this property (for example completely distributive, or 5-element nondistributive). Here is a small counterexample.

Let $L$ be the 9-element lattice $$L=\{s_1,t_1,s^*,t^*,s_1\wedge t_1,s_1\wedge t^*, s^*\wedge t_1,0,1\}$$ where all elements are incomparable except that $0$, $1$, $\wedge$ have the usual meaning. Then $S=\{s_1,s^*\}$ and $T=\{t_1,t^*\}$ are both mutually-disjoint, but $S\wedge T$ is not, since $$0< s_1\wedge t_1=(s_1\wedge t_1)\wedge \left[ (s_1\wedge t^*)\vee (s^*\wedge t_1)\right].$$

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How are all elements incomparable? $s_1 \wedge t_1$ and $s_1$ are comparable. In your example, $S \wedge T$ consists of those three joins, and 0, and this set is certainly mutually disjoint. –  Amit Kumar Gupta Nov 12 '10 at 10:03
Oh sorry, I missed the $\wedge$ in "0, 1, $\wedge$ have the usual meaning." Never mind anything in my previous comment. –  Amit Kumar Gupta Nov 12 '10 at 19:58

Here's an answer to a related question which involves much more standard terminology.

Say $A \subset L$ is an antichain iff $\forall x, y \in A(x \neq y \rightarrow x \wedge y = 0)$. If $A_1$ and $A_2$ are two antichains, then $A_1 \wedge A_2$ is yet another antichain which we say "refines" both $A_1$ and $A_2$.

Proof: If $x_i, y_i \in A_i$ with $x_i \wedge y_i = 0$, then clearly $(x_1 \wedge x_2) \wedge (y_1 \wedge y_2) = 0. \\ \\ \square$

The notion of antichain makes sense even if the joins $\bigvee (S - \{ x \})$ aren't defined. If $L$ satisfies $\forall x \in L\ \forall S \subset L\ (x \wedge \bigvee S = \bigvee (\{ x \} \wedge S)$ then the notions of "antichain" and "mutually disjoint" coincide.

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I think it is more relevant to calculate $(x_1\wedge x_2)\wedge(y_1\wedge y_2)=0$ in your proof, since it is $x_1\wedge x_2$ that is in $A_1\wedge A_2$, etc. (although I understand the expressions are equivalent). –  Joel David Hamkins Nov 12 '10 at 13:26
Yeah, that's what I meant; fixed it. –  Amit Kumar Gupta Nov 12 '10 at 20:00