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Let $X$ be an ordered set. A down-set (also called a lower set or an order ideal) of $X$ is a subset $D$ of $X$ such that for every $x, y \in D$, if $x \in D$ and $y \leq_X x$, then $y \in D$. The down-set lattice $\mathcal{O}(X)$ of $X$ is the set of all down-sets of $X$ ordered by inclusion.

Let $P$ and $Q$ be ordered sets, and let $\mathcal{O}(P)$ and $\mathcal{O}(Q)$ be their respective down-set lattices. Are $P$ and $Q$ order-isomorphic if and only if $\mathcal{O}(P)$ and $\mathcal{O}(Q)$ are order-isomorphic? If $P$ and $Q$ are order-isomorphic, then it's easy to show that $\mathcal{O}(P)$ and $\mathcal{O}(Q)$ are order-isomorphic. And if $P$ and $Q$ are totally ordered, then $\mathcal{O}(P)$ and $\mathcal{O}(Q)$ are order-isomorphic to the ideal completions of $P$ and $Q$ respectively, and it's again easy to show that if $\mathcal{O}(P)$ and $\mathcal{O}(Q)$ are order-isomorphic, then $P$ and $Q$ are order-isomorphic. But what if $P$ and $Q$ are only partially ordered?

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2 Answers 2

up vote 1 down vote accepted

If $\mathcal{O}(P)$ and $\mathcal{O}(Q)$ are order-isomorphic, then $P$ and $Q$ are isomorphic as well. Furthermore, not only can we recover the poset $P$ from the lattice $\mathcal{O}(P)$, but the mapping $P\mapsto\mathcal{O}(P)$ gives us a duality between the category of all posets and certain kinds of lattices called superalgebraic lattices.

If $L$ is a complete lattice, then an element $x\in L$ is said to be supercompact if $x\leq\bigvee R$ implies that $x\leq r$ for some $r\in R$. A complete lattice $L$ is said to be superalgebraic if every element in $L$ is the least upper bound of supercompact elements.

In the lattice $\mathcal{O}(P)$ it is easily seen that the supercompact elements are precisely elements of the form $\downarrow x=\{y\in P|y\leq x\}$. Therefore the original poset $P$ is order isomorphic to the set of all supercompact elements in $\mathcal{O}(P)$. Furthermore, it is easily seen that $\mathcal{O}(P)$ is a superalgebraic lattice since every lower set is the union of sets of the form $\downarrow x$.

For more information on superalgebraic lattices and this duality see for example the paper Complete Congruences on Topologies and Down-set Lattices by Erne, Gehrke, and Pultr.

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I wasn't aware of the previous work which called them "superalgebraic lattices". In my book on Lipschitz algebras I called them "Stone lattices" because they exhibit some analogies to Stone spaces. (In this analogy, completely distributive complete lattices correspond to compact Hausdorff spaces.)

To add to Joseph's answer, you can identify ${\mathcal O}(P)$ with the set of order-preserving maps from $P$ into the two-element lattice ${\bf 2}$. And you recover $P$ as the set of complete $0,1$-lattice homomorphisms from ${\mathcal O}(P)$ into ${\bf 2}$.

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Concretely, a complete $0,$1-lattice homomorphism from ${\mathcal O}(P)$ into ${\bf 2}$ corresponds to a pair of elements $x,y$ such that ${\mathcal O}(P)$ is the disjoint union of $x\downarrow$ and $y\uparrow$ (the set of elements $\leq x$ and the set of elements $\geq y$, respectively). We have $x=\sup f^{-1}(0)$ and $y=\inf f^{-1}(1)$. –  Nik Weaver Apr 1 '13 at 4:00

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