free bounded lattice

Question

Suppose I have a partial order and I want to construct the free bounded lattice that contains that partial order. Suppose our PO has $A<X$, $B<X$, $A<Y$ and $B<Y$, with no intermediate positions. Join(A,X) is going to be X and Meet(A,X) is going to be A. We need to add in Join(A,B) which is above A and B, and below X and Y. Suppose we've added in all needed extra Joins for any subset of elements in our PO. It seems that we don't need to add any more elements to make our free lattice. To find the Meet of a subset we can just look at the set of elements in our PO that are below all elements in our set and take the Join of that to be our Meet. (With all elements in the PO being below the empty set of elements, to make our bounded lattice work -- which makes sense because the Meet gets bigger the less elements there are to meet.) Is that correct?

Answer 1

The construction you describe does produce a bounded lattice, but it does not not produce the bounded lattice freely generated by the poset.

One way to define the bounded lattice freely generated by the poset ${\bf P}=\langle P; \leq \rangle$ is the bounded lattice $L({\bf P})$ with the presentation $$ \langle P\;|\;\{(a,a\wedge b)\;|\;a\leq b\}\rangle. $$ This is a presentation relative to the variety of bounded lattices.

$L({\bf P})$ has the property that any order-preserving function $f\colon {\bf P}\to B$ from ${\bf P}$ to a bounded lattice $B$ extends uniquely to a bounded lattice homomorphism $\widehat{f}\colon L({\bf P})\to B$.

As I claimed in the first paragraph, the construction in the question, when applied to a finite poset, produces a bounded lattice, but it is not free as a bounded lattice generated by ${\bf P}$. You can see this from the fact that when the construction is applied to a finite poset it always produces a finite bounded lattice, while the free bounded lattice generated by a poset is almost never finite. The specific poset in the question is called the $2$-crown by some folks. The construction in the question starts with a $2$-crown and constructs a $7$-element lattice. The free bounded lattice generated by a $2$-crown has size $8$. If we had started with two independent chains instead of a $2$-crown (say, $A<X$ and $B<Y$ with no other relations), then the construction of the question would produce a $9$-element bounded lattice, while the free bounded lattice generated by $2$ independent chains is infinite. (This latter fact is a theorem of Howard Rolf from 1958.)

The main paper that is relevant to this question is

R. A. Dean, Sublattices of free lattices, in Lattice Theory Amer. Math. Soc., Providence, Rhode Island, 1961, Proc. Symp. Pure Math. II, pp. 31–42.

while a modern treatment can be found in the book

Ralph Freese, Jaroslav Jezek and J. B. Nation, Free Lattices, Amer. Math. Soc. (Mathematical Surveys and Monographs 42), 1995.

Answer 2

A different approach from the one in Keith Kearnes' answer is following the definition of free lattice generated by a poset, in Georege Grätzer's, General Lattice Theory, Ch.I, section 5.
(Actually he defines it by two different, apparently non-equivalent ways: Definition 2 and Definition 23. I don't know if the definition adopted by Kearnes is equivalent to one of these; it seems to give the same result in this example, judging by the number of elements.)

Definition 2. Let $P$ be a poset and let $\mathbf K$ be a variety of lattices. A lattice $F_{\mathbf K}(P)$ is called a free lattice over $\mathbf K$ generated by $P$ iff the following conditions are satisfied:

(1) $F_{\mathbf K}(P) \in \mathbf K$.

(2) $P \subseteq F_{\mathbf K}(P)$, and for $a,b,c \in P$, $\inf\{a,b\}=c$, in $P$ iff $a\wedge b=c$ in $F_{\mathbf K}(P)$, and $\sup\{a,b\}=c$, in $P$ iff $a\vee b=c$ in $F_{\mathbf K}(P)$.

(3) $[P] = F_{\mathbf K}(P)$

(4) Let $L \in \mathbf K$ and let $\varphi:P\to L$ be an isotone map with the following two properties, for $a,b,c \in P$:
     (4.1) if $\inf\{a,b\}=c$ in $P$, then $\varphi(a)\wedge \varphi(b)=\varphi(c)$ in $L$,
     (4.2) if $\sup\{a,b\}=c$ in $P$, then $\varphi(a)\vee \varphi(b)=\varphi(c)$ in $L$.
Then there exists a (lattice) homomorphism $\psi:F_{\mathbf K}(P)\to L$ extending $\varphi$ (that is, satisfying $\varphi(a)=\psi(a)$, for all $a\in P$).

Here, given a lattice $L$ and $A \subseteq L$, $[A]$ is the sub-lattice of $L$ generated by $A$.

In this case, it turns out to be simple to check the conditions, since the only subsets of $P$ for which there exists infimum or supremum are the ones which form chains.
Also, in this case, $\mathbf K$ is the class of bounded lattices.

It is easy to verify that the free bounded lattice generated by your poset $P=\{a,b,x,y\}$, with $a<x$, $a<y$, $b<x$ and $b<y$ is the following

Apparently you were thinking that $a\vee b = x \wedge y$.
The lattice that differs from the one above by that condition satisfies conditions (1), (2) and (3) in the definition, but not (4).
Indeed, if we call $F(P)$ to that 7-element lattice and $L$ to the one in the picture, with $\varphi$ the inclusion map, we see that there exists no $\psi$ extending $\varphi$, since we would have $a\vee b=x\wedge y$, in $F(P)$, and thus, it should be $\psi(a\vee b) = \psi(x\wedge y)$, which doesn't happen in $L$.