Let \(p \in P\) be an element in a preorder. Consider \(A = \{p\}\)
Show that \(\wedge A \cong p\)
Show that if \(P\) is a partial order, then \(\wedge A = p\)
Are the analogous facts true when \(\wedge\) is replaced by \(\vee\)?
The first condition of the meet gives us that \(\wedge A \leq p\).
The second condition is that \(\forall q \in P: q \leq p \implies q \leq \wedge A\).
Substituting \(p\) in for \(q\), the antecedent holds such that we get \(p \leq \wedge A\).
Therefore \(p \cong \wedge A\)
The difference between a partial order and a preorder is that congruent elements are equal, so we directly get that \(p = \wedge A\)
Yes, the argument is perfectly symmetric.