There could be multiple meets/joins, but the definition forces them to be isomorphic.
An arbitrary preorder need not have a meet nor join.
E.g a two element discrete preorder has no overall meet/join, because the meet must be less/greater than or equal to both elements in the set.
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.
For a preorder \((P, \leq)\), the meet and join of \(A \subseteq P\).
The meet \(\wedge A\) is an element such that
\(\forall a \in A: \wedge A \leq a\)
\(\forall q \in P: (\forall a \in A: q \leq a) \implies q \leq \wedge A\)
Think of as a GREATEST LOWER BOUND
The join \(\vee A\) is an element such that
\(\forall a \in A: a \leq \vee A\)
\(\forall q \in P: (\forall a \in A: a \leq q) \implies \vee A \leq q\)
Think of as a LEAST UPPER BOUND
In a total order, the meet of a set is its infimum, while the join is the supremum.
Note that \(\mathbb{B}\) is a total order, to generalize Example 1.88.
Suppose \((P,\leq)\) is a preorder and \(A \subseteq B \subseteq P\) are subsets that have meets. Then \(\bigwedge B \leq \bigwedge A\)
Let \(m = \bigwedge A\) and \(n = \bigwedge B\).
For any \(a \in A\) we also have \(a \in B\), so \(n \leq A\) because \(n\) is a lower bound for \(B\).
Thus \(n\) is also a lower bound for \(A\) and hence \(n \leq m\) because \(m\) is \(A\)’s greatest lower bound.