But it contains only $u$ and $v$, so $uv$ is an edge.

Some Examples {#sect:examples}
=============

As all but the minimum and maximum vertex of a lattice must have at least one cover and be covered by one other vertex, the following is immediate from Proposition \[prop:HasseSubgraph\].

\[ex:degOne\] For a connected reflexive graph $G$ with a degree one vertex $v$, $v$ must be the minimum or maximum vertex of any compatible lattice $L$. In particular, the only reflexive trees with compatible lattices are paths.

(1,0.43333335)(-.2,0) (0,0)[![Graph (left) with compatible lattice (right) but no compatible distributive lattice[]{data-label="fig:LnotDL"}](./LnotDL.pdf "fig:"){width="\unitlength"}]{} (0.46666669,0.02500007)[(0,0)\[lb\]]{} (0.46666669,0.36666671)[(0,0)\[lb\]]{} (0.45833336,0.10833334)[(0,0)\[lb\]]{} (0.38553573,0.14279105)[(0,0)\[lb\]]{} (0.49388894,0.14195102)[(0,0)\[lb\]]{} (0.45693121,0.18394846)[(0,0)\[lb\]]{} (0.45273149,0.23854504)[(0,0)\[lb\]]{} (0.34437829,0.19822752)[(0,0)\[lb\]]{} (0.54596561,0.20326726)[(0,0)\[lb\]]{} (0.3905754,0.26206356)[(0,0)\[lb\]]{} (0.50396828,0.25366398)[(0,0)\[lb\]]{} (0.4636508,0.29230159)[(0,0)\[lb\]]{}

\[prop:badretracts\] Neither the class of graphs admitting compatible lattices, nor the class admitting compatible distributive lattices, are closed under retraction.

It is easy to see that the reflexive biclique $K_{1,4}$ is a retract of the product $P_2^2$ of two reflexive paths. $P_2^2$ has a distributive lattice by Lemma \[lem:product\], but $K_{1,4}$ does not, by Example \[ex:degOne\].

\[ex:bigone\] There are graphs that have compatible lattices but have no compatible distributive lattices.

Let $G$ be the graph on the left of Figure \[fig:LnotDL\]. It is easy but tedious to verify, using that the 