Smallest lattice containing a poset

Question

Given a poset $P$(we can assume $P$ is finite if necessary), how can we construct the smallest lattice containing $P$? (Does this exist?) To make the question precise, I am looking for a lattice $L$ and an order-preserving inclusion $\iota : P \to L$ which satisfies the following universal property:

If $L'$ is a lattice and if $f : P \hookrightarrow L' $ is an order-preserving injection, then there exists a unique lattice morphism $\varphi : L \to L'$ such that $\varphi \circ \iota = f$

Answer 1

Yes, this is possible - but it is incredibly difficult to describe explicitly.

First, the really general way to do this is to use the fact that lattices are algebraic structures and to basically write a presentation of the lattice you want, then just compose the universal properties of freeness and quotients. In particular, if $(P,\leq)$ is your poset, you can consider the free lattice on $P$ modded out by the relation $a \vee b = b$ for each pair $a\leq b$ in $P$. Call that lattice $L$.

There is then an obvious order-preserving map $\iota:P\rightarrow L$ which has the universal property that for any $f:P\rightarrow L'$ into another lattice, there is a unique lattice morphism $\varphi:L\rightarrow L'$ such that $f=\varphi\circ \iota$. You can check that $\iota$ is actually injective by using this universal property on the map $f$ taking $P$ into the powerset of $P$ by the rule $$f(x)=\{z\in P: x\geq z\}$$ and noting that this $f$ is injective, so $\iota$ must also be.

Unfortunately, this brings us to the bad news: It is hard to describe even what a free lattice on three elements looks like (it is, for one thing, infinite) - but, of course, if we apply this construction to a poset on three incomparable elements, that's exactly what we get. All this construction says is "the lattice consists of the set of expressions using the operations $\vee$ and $\wedge$ on terms which are either themselves expressions or terms from $P$ modulo an equivalence relation generated by the lattice axioms and the relations $a\vee b \sim b$."

There's also another nice general construction that somehow feels even less explicit, but is a bit easier to verify without any prior knowledge:

Let $P$ be a poset and $S$ be a sufficiently large set*. Let $F$ be the set of all tuples $(\vee, \wedge, \iota)$ such that $(S,\vee,\wedge)$ is a lattice and $\iota:P\rightarrow S$ is an order preserving map into $(S,\vee,\wedge)$. Then, consider the lattice $$\mathscr L = \prod_{(\vee, \wedge, \iota)\in F}(S,\vee,\wedge)$$ and the map $\iota : P\rightarrow\mathscr L$ given by the product of the maps $\iota$ to each factor. The lattice generated by the image of $\iota$ satisfies the universal property.

(*Any set at least as large as every lattice generated by a set of size $|P|$ would suffice; assuming the axiom of choice, this means "countable" when $P$ is finite, and just "equally large as $P$" when $P$ is infinite)

Answer 2

I think you're looking for the Dedekind–MacNeille completion. To paraphrase Theorem 5.3.8 in the book Ordered Sets by Bernd Schroeder, the Dedekind–MacNeille completion is the smallest complete lattice $L$ containing $P$, i.e. there is an embedding $\iota : P \to L$ and for any other complete lattice $L'$ with an embedding $f : P \to L'$ there is an embedding $\phi : L \to L'$ such that $f = \phi \circ \iota$. Since every finite lattice is complete, for finite lattices the Dedekind–MacNeille completion is simply the smallest lattice.

"Embedding" refers to an order embedding, an injective function for which $a \leq b$ iff $f(a) \leq f(b)$, i.e. it is both order-preserving and order-reflecting. So compared to your precise question it is a stronger notion than the "order-preserving injection" but weaker than the lattice (homo)morphism. But the Dedekind–MacNeille completion of $P$ does preserve all joins and meets of $P$.