Is every partial order a suborder of a lattice?

Question

Is every partial order a suborder of a lattice? If not, can someone give me a counterexample, preferably a finite one if it exists.

Answer 1

Given a poset $\mathbb{P}$, let $Down(\mathbb{P})$ be the set of downwards-closed subsets of $\mathbb{P}$. Now $Down(\mathbb{P})$ is itself a poset ordered by inclusion, and there is an embedding $$i: \mathbb{P}\rightarrow Down(\mathbb{P}): p\mapsto \{q: q\le p\}.$$

The fact that $Down(\mathbb{P})$ is a lattice follows from two simple observations:

• $Down(\mathbb{P})$ is closed under $\cup$ and $\cap$.

• Any family of sets closed under $\cup$ and $\cap$, partially ordered by $\subseteq$, is a lattice (with join and meet given by $\cup$ and $\cap$ respectively).

In fact, $Down(\mathbb{P})$ is always a complete distributive lattice.

(It's worth noting that this is very close to what Berci mentioned in their comment above: their idea corresponds to the sublattice of $Down(\mathbb{P})$ generated by the image of $i$.)

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It's worth noting however that this construction is quite ugly. For example, it will add elements even when none are needed: the least upper bound of $i(a)$ and $i(b)$ in $Down(\mathbb{P})$ is always $\{c: c\le a\wedge c\le b\}$, which is not in the image of $i$ if $a$ and $b$ are incomparable in $\mathbb{P}$. However, it does do the job.