According to the Wikipedia article for 'Lattice', a sublattice is a non-empty subset of a lattice $L$ that is a lattice with the same meet and join operations as $L$.

I'd like to know if there is a name for the following generalization of this concept. Let $\mathbf{P}=(P,\leq)$ be a poset (which is not necessarily a lattice), and let $L\subseteq P$ satisfy the following conditions.

  1. For every $x,y \in L$ there is a greatest lower bound for $\{x,y\}$ in $\mathbf{P}$, and it belongs to $L$.
  2. For every $x,y \in L$ there is a least upper bound for $\{x,y\}$ in $\mathbf{P}$, and it belongs to $L$.

Is there a standard name for the type of "sub-lattice" that $L$ is?

It is technically not a sublattice, because $\mathbf{P}$ is not known to be a lattice. However, it won't do simply to say that $(L,\leq)$ forms a lattice, because it is possible for a subset $L$ of $P$ to form a lattice under $\mathbf{P}$'s partial order without satisfying the above conditions. Take, for instance, the following counter-example.

Denote by $P$ the powerset of $\{0,1,2\}$, and denote by $\leq$ set containment. Set $L := \left\{\emptyset, \{0\},\{1\}, P\right\}$. Then $(L,\leq)$ is a lattice, however it doesn't satisfy condition 2 above, because the least upper bound for $\left\{\{0\},\{1\}\right\}$ in $\mathbf{P} = (P,\leq)$ is $\{0,1\} \notin L$.


The same type of question arises frequently in semigroup theory. Let $M$ be a monoid with identity $1$ and let $e \not= 1$ by an idempotent of $M$. You may have a subsemigroup of $M$ which forms a monoid (or even a group) with identity $e$. For instance the subsemigroup $eMe$ is a monoid with identity $e$ and the $\mathcal{H}$-class of $e$ is a group.

In semigroup theory, various names have be proposed. The monoid $eMe$ is sometimes called the local monoid at $e$. A subsemigroup of a semigroup $S$ which turns out to be a group is sometimes called a group in $S$ (or inside $S$).

To come back to your question, you may call $L$ an ordered subset forming a lattice or perhaps a lattice inside $\mathbf{P}$. But don't forget to give a precise definition if you want to use this non standard terminology.

  • $\begingroup$ Thanks. This is interesting, however, I'm not sure the analogy is quite correct, because as my counterexample shows, it is possible for a subset $L$ of $P$ to form a lattice in $\mathbf{P}$ without satisfying the two "closure" requirements I listed. I think that your terminology is more suitable for describing the $L$ in my counterexample. $\endgroup$ – Evan Aad Jun 11 '18 at 13:57
  • $\begingroup$ You are perfectly right. I will delete my answer within a few minutes. $\endgroup$ – J.-E. Pin Jun 11 '18 at 14:21
  • $\begingroup$ I actually find the analogy to be correct, but maybe I'm missing something. $\endgroup$ – amrsa Jun 11 '18 at 14:24
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    $\begingroup$ @amrsa Yes, that's right. But the point is that if you start with a monoid with identity $1$, and a group $G$ inside $M$ with identity $e$, then $1$ also acts as an identity for the elements of $G$, but does not belong to $G$. Thus you cannot say "if there is an identity for $G$ in $M$, then this identity belongs to $G$". $\endgroup$ – J.-E. Pin Jun 11 '18 at 14:38
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    $\begingroup$ @amrsa. OK, thank you, I will not delete my answer. I think the comments now clearly explain the difficulty of the question. $\endgroup$ – J.-E. Pin Jun 11 '18 at 14:55

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