Let $(X,\leq)$ be a lattice.
$(X,\leq)$ is said to be distributive when for any $x$, $y$ and $z$ in $X$ it holds that
\begin{equation*}
x\wedge(y\vee z) = (x\wedge y)\vee(x\wedge z),
\end{equation*}
or equivalently
\begin{equation*}
x\vee(y\wedge z) = (x\vee y)\wedge(x\vee z).
\end{equation*}
$(X,\leq)$ is said to be distributive as a meet-semilattice
when for any $x$, $y$ and $z$ in $X$,
if $y\wedge z\leq x$ then there are $y'$ and $z'$ in $X$
such that $y\leq y'$ and $z\leq z'$ and $y'\wedge z' = x$.
If $(X,\leq)$ is distributive as a lattice,
then for any $x$, $y$ and $z$ in $X$
satisfying $y\wedge z\leq x$ it holds that
\begin{equation*}
x = x\vee(y\wedge z) = (x\vee y)\wedge(x\vee z)
\end{equation*}
and $y\leq x\vee y$ and $z\leq x\vee z$,
so $(X,\leq)$ is distributive as a meet-semilattice.
According to
Wikipedia
the converse implication also holds true:
if $(X,\leq)$ is distributive as a meet-semilattice
then it is a distributive lattice.
My question: How to show that
if $(X,\leq)$ is distributive as a meet-semilattice
then it is a distributive lattice?
Best Answer
If it is not a distributive lattice, then it has a sub-lattice isomorphic either to the diamond $M_3$ or the pentagon $N_5$.
It's easy to check that neither of these is distributive as a semi-lattice.
In the case of $M_3$, just take $x,y,z$ to be the atoms; in the case of $N_5$, take, for example $$0\prec x \prec z \prec 1 \quad\text{and}\quad 0 \prec y \prec 1.$$
While all of the above is true, it doesn't prove that a distributive semilattice which is also a lattice is a distributive lattice.
The reason is that the condition that defines distributive semilattices is not equational (or quasi-equational for that matter) and so, for the best of my knowledge, distributive semilattices don't form a class closed for the formation of subalgebras.
It turns out, it's not difficult to produce a direct proof.
So suppose that a lattice $L$ satisfies the condition $$x\leq y \vee z \implies \exists y',z' \left( y'\leq y,\, z'\leq z,\, x=y'\vee z' \right).$$ Now we want to prove that $L$ satisfies $$x\land (y\lor z)=(x\land y)\lor(x\land z).$$ Certainly $L$ (as all lattices) satisfy $$x\land (y\lor z)\geq(x\land y)\lor(x\land z),$$ so we have to prove the converse. But $$x\land (y \lor z) \leq y \lor z,$$ and so there exist $y'\leq y$ and $z'\leq z$ such that $$x\land(y\lor z)=y'\lor z'.$$ But then it follows that $x\geq y'\vee z'$, and so $x\geq y'$ and $x\geq z'$, yielding $$y'=x\wedge y' \quad\text{and}\quad z'=x\wedge z'.$$ Hence $$x\land(y\lor z)=(x\land y')\lor(x\land z')\leq (x\land y)\lor (x\land z).$$