Let \(S=\{1,2,3\}\)
Come up with any preorder relation on \(S\), and define \(U(\leq):=\{(s_1,s_2)\ |\ s_1 \leq s_2\}\) (the relation ‘underlying’ the preorder. Note \(\mathbf{Pos}(S) \xhookrightarrow{U} \mathbf{Rel}(S)\))
Pick binary relations such that \(Q \subseteq U(\leq)\) and \(Q' \not \subseteq U(\leq)\)
We want to check that the reflexive/transitive closure operation \(Cl\) is really left adjoint to the underlying relation \(U\).
The meaning of \(Cl \dashv U\) is \(Cl(R) \sqsubset \leq \iff R \sqsubset U(\leq)\), where \(\sqsubset\) is the order relation on \(\mathbf{Pos}(S)\)
Concretely show that \(Cl(Q) \sqsubset \leq\)
Concretely show that \(Cl(Q') \not \sqsubset \leq\)
Let the preorder be given by this diagram (with implicit reflexive arrows):
Let \(Q\) be given by the following diagram
and let \(Q'=S\times S\)
\(Cl(Q) = \{11,12,22,33\}\) \(\sqsubset\) \(\leq = \{11,22,33,12,23,13\}\)
\(Cl(Q') = Q' = S \times S\) \(\not \sqsubset\) \(\leq\) (reason: \((3,1) \in S \times S\) but \((3,1) \not \in \leq\))