Show that the following are directed sets:
Any simply ordered set, under the relation \(\leq\).
The collection of all subsets of a set \(S\), partially ordered by inclusion (that is, \(A \preceq B\) if \(A \subset B\)).
A collection \(\mathcal{A}\) of subsets of \(S\) that is closed under finite intersections, partially ordered by reverse inclusion (that is, \(A \preceq B\) if \(A \supset B\)).
The collection of all closed subsets of a space \(X\), partially ordered by inclusion.
Solution
For a directed set, every pair of indices must have a common upper bound.
In a simply ordered set, for \(a,b\), whichever is larger is a common upper bound.
For subsets \(A,B\subset S\) ordered by inclusion, \(A\cup B\) is a common upper bound.
If \(\mathcal A\) is closed under finite intersections and ordered by reverse inclusion, then \(A\cap B\in\mathcal A\), and because \(A\supset A\cap B\) and \(B\supset A\cap B\), we have \(A\preceq A\cap B\) and \(B\preceq A\cap B\).
Closed subsets of \(X\), ordered by inclusion, have \(A\cup B\) as a common upper bound because finite unions of closed sets are closed.