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.