2 tried to explain why the current answers are still insufficient

For a given poset, (I think that) it is easy to construct the minimal join-semilattice containing that poset. I wonder whether the minimal lattice containing that poset is also easy to construct. I can't even prove that it exists, but this is probably only due to the fact that I didn't specify what I mean by "minimal" here (because I don't know which definition would work for a lattice).

By "it is easy to construct the minimal join-semilattice...", I mean the following construction:

Let $P$ be a partially ordered set. For $X\subset P$, let $X^{u} := \{ p\in P : x \leq p \;\forall x \in X \}$. Let $J(P):=\{X\subset P:1\leq |X| < \infty\}$ and define a preorder $\leq_u$ on $J(P)$ via $X\leq_u Y$ iff $Y^u \subset X^u$. (It is easy to check that $\leq_u$ is reflexive and transitive.) Denote the quotient of $J(P)$ by the equivalence relation $\equiv_u$ associated to $\leq_u$ by $S:=J(P)/\!\!\equiv_u$ and let $j:J(P)\mapsto S$ denote the canonical projection to $S$. Then $S$ is a join-semilattice with $j(X)\lor j(Y)=j(X\cup Y)$, because we have $(X\cup Y)^u=X^u\cap Y^u$. I think I could also prove that it is the minimal join-semilattice containing (an isomorphic copy of) the poset $P$.

Edit The current answers and comments basically propose to start with any lattice containing the given poset, and then take the sublattice generated by the elements of the poset. Below, I sketched two lattices to explain why the problem is more complicated than that. The lattice on the left contains an order-embedding of the lattice on the right, but the sublattice of the left lattice generated by the elements of the right lattice is not isomorphic to the right lattice (because the meet of A and B is M instead of m).

  1        1
/ \      / \
A   B    A   B
\ /      \ /
M
|        m
m
/ \      / \
a   b    a   b
\ /      \ /
0        0


The above example is no counterexample to the "minimal" embedding using the Dedekind-MacNeille completion as starting point (as suggested by Joseph van Name). However, it should clarify that it isn't obvious that this construction really gives the minimal lattice.

1

# Minimal (semi)lattice containing a given poset

For a given poset, (I think that) it is easy to construct the minimal join-semilattice containing that poset. I wonder whether the minimal lattice containing that poset is also easy to construct. I can't even prove that it exists, but this is probably only due to the fact that I didn't specify what I mean by "minimal" here (because I don't know which definition would work for a lattice).

By "it is easy to construct the minimal join-semilattice...", I mean the following construction:

Let $P$ be a partially ordered set. For $X\subset P$, let $X^{u} := \{ p\in P : x \leq p \;\forall x \in X \}$. Let $J(P):=\{X\subset P:1\leq |X| < \infty\}$ and define a preorder $\leq_u$ on $J(P)$ via $X\leq_u Y$ iff $Y^u \subset X^u$. (It is easy to check that $\leq_u$ is reflexive and transitive.) Denote the quotient of $J(P)$ by the equivalence relation $\equiv_u$ associated to $\leq_u$ by $S:=J(P)/\!\!\equiv_u$ and let $j:J(P)\mapsto S$ denote the canonical projection to $S$. Then $S$ is a join-semilattice with $j(X)\lor j(Y)=j(X\cup Y)$, because we have $(X\cup Y)^u=X^u\cap Y^u$. I think I could also prove that it is the minimal join-semilattice containing (an isomorphic copy of) the poset $P$.