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Let $(L,\vee,\wedge,0,1)$ be a lattice with unique least and greatest elements $0$ and $1$, respectively. I'll say that an antichain $A$ in $L$ is a subset of $L\setminus\{0\}$ such that for every $a,b\in A$ distinct, $a\wedge b=0$. (This is a set-theorist's antichain.)

I want to consider the following "finite-join" variation on an antichain: a subset $A$ of $L\setminus\{0\}$ has property (A) if for all $a_1,\ldots,a_n,b_1,\ldots,b_m\in A$ distinct, $(a_1\vee\cdots \vee a_n)\wedge(b_1\vee\cdots\vee b_n)=0$.

Note that in a distributive lattice, (A) is equivalent to being an antichain.

To see that they're not equivalent, consider the lattice of subspaces of a vector space of dimension 6 with basis vectors $e_1,e_2,e_3,e_4,e_5,e_6$. Let $a$ be the subspace generated by $(e_1,e_4)$, $b_1$ the subspace generated by $(e_2+e_3,e_5+e_6)$, and $b_2$ the subspace generated by $(e_1+e_2+e_3,e_4+e_5+e_6)$. Then, $a\wedge b_1=a\wedge b_2=\{0\}$, but $a\wedge(b_1\vee b_2)=a$.

My questions: 1. Does property (A) have a name in the literature? Is it a studied notion?

  1. When does (A) coincide with being an antichain? Only in distributive lattices?

  2. Does maximal with respect to (A) imply being a maximal antichain?

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  • $\begingroup$ Do you want to pick n and m in advance, and then define property A? The sub universe of a's and b's under join will have to avoid 1 in order for your property to hold for varying n and m. Gerhard "Is Rusty On Lattice Theory" Paseman, 2016.10.22. $\endgroup$ – Gerhard Paseman Oct 23 '16 at 3:41
  • $\begingroup$ I want n and m to vary. Typically I am thinking about infinite (even uncountable) lattices. $\endgroup$ – Iian Smythe Oct 23 '16 at 3:43
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I will add a few comments to Bjørn Kjos-Hanssen's answer.

  1. Does property (A) have a name in the literature? Is it a studied notion?

The property is a kind of independence property. A set-theorist's antichain $S$ of a lattice $L$ is called weakly independent if whenever $a_1,\ldots,a_{n+1}\in S$ are distinct, then $(a_1\vee\cdots\vee a_n)\wedge a_{n+1} = 0$. It is called strongly independent if whenever $I$ and $J$ are finite sets of indices, then $(\bigvee_{i\in I} a_i)\wedge (\bigvee_{j\in I} a_j)=(\bigvee_{k\in I\cap J} a_k).$ Property (A) lies between these two concepts. In a modular lattice all three notions agree (weak independence, Property (A), and strong independence).

  1. When does (A) coincide with being an antichain? Only in distributive lattices?

It coincides in any pseudocomplemented lattice. These need not be distributive. For example, the congruence lattice of a semilattice is pseudocomplemented and almost never distributive. The order-dual of the lattice of convex subsets of a finite chain is pseudocomplemented, but not distributive when the chain has more than 2 elements.

  1. Does maximal with respect to (A) imply being a maximal antichain?

I would guess that being maximal with respect to (A) rarely means being a maximal antichain for lattices that are not pseudocomplemented. For a simple example, if $S$ is an antichain in the lattice $L$ of subspaces of a vector space $V$, and $S$ satisfies Property (A), then the size of $S$ cannot exceed the dimension of $V$. But $L$ can have antichains larger than the dimension of $V$. For example, if $V$ has finite dimension greater than one and $T\subseteq L$ is the set of 1-dimensional subspaces, then $T$ is larger than any antichain in $L$ that has Property (A).

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  • $\begingroup$ A more comprehensive answer than I could have hoped for. Thank you! $\endgroup$ – Iian Smythe Oct 23 '16 at 19:04
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  1. Does property (A) have a name in the literature? Is it a studied notion?

Don't know, but we could call it the property of being a strong antichain (since by taking $m=n=1$, it implies being an antichain).

  1. When does (A) coincide with being an antichain? ...

I don't have a general answer...

... Only in distributive lattices?

No. It coincides in any lattice all of whose antichains have size $\le 2$, and these can be non-distributive. Example: the non-modular (hence non-distributive) 5-element lattice $N=\{0,a,b,c,1\}$ where $0<a<1$, $0<b<c<1$, and $a$ is incomparable with $b$ and $c$.

  1. Does maximal with respect to (A) imply being a maximal antichain?

No, consider the 5-element modular non-disitributive lattice $M=\{0,a_1,a_2,a_3,1\}$ with $0<a_i<1$ for each $i$, and the $a_i$ incomparable. Then $\{a_1,a_2\}$ is a maximal strong antichain, but not a maximal antichain since $\{a_1,a_2,a_3\}$ is an antichain and $a_1\wedge (a_2\vee a_3)=a_1\wedge 1 = a_1\ne 0$.

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