Way back in Lecture 6 we saw how a right adjoint for \\(f\\), if it exists, has to send \\(b\\) to the sup of \\(G(b) = \\{a \in A : \; f(a) \le b \\}\\).

So it is necessary and sufficient for \\(A\\) to have all sups of sets of form \\(G(b)\\).

It strikes me that \\(G(b)\\) is a "down-set": \\(a'' \le a' \in G(b) \implies f(a'') \le f(a') \le b \implies a'' \in G(b)\\)

The down-sets of \\(A\\) form a topology on \\(A\\) – if that topology is compact then \\(G(b)\\) is the union of a finite number of basic downsets of form \\(\\{ a \in A : a \le a_i \\}\\), and then the sup of \\(G(b)\\) would be the join of the \\(a_i\\). Not sure how useful this is tho.

So it is necessary and sufficient for \\(A\\) to have all sups of sets of form \\(G(b)\\).

It strikes me that \\(G(b)\\) is a "down-set": \\(a'' \le a' \in G(b) \implies f(a'') \le f(a') \le b \implies a'' \in G(b)\\)

The down-sets of \\(A\\) form a topology on \\(A\\) – if that topology is compact then \\(G(b)\\) is the union of a finite number of basic downsets of form \\(\\{ a \in A : a \le a_i \\}\\), and then the sup of \\(G(b)\\) would be the join of the \\(a_i\\). Not sure how useful this is tho.