John, continuing the discussion from Lecture 21, I became interested in the following, maybe you'll be able to help me here. Let's say we have a monotone function \\(f: X \times X \to X\\) and the condition \\(C^\times: x \le x' \textrm{ and } y \le y' \Rightarrow f(x, y) \le f(x', y') .\\)

As we learned, the condition implies that the function is monotone: \\(C^\times \Rightarrow f \in M^\times\\), where \\(M^\times\\) is the set of monotone functions \\(X \times X \to X\\). The question is whether converse \\(f \in M^\times \Rightarrow C^\times\\) is also true, that is the function \\(f\\) being monotonic implies that it supports the condition \\(C^\times\\)?

The easiest way to show that converse is not true, is to find a monotone function \\(g \in M^\times\\) which doesn't support \\(C^\times\\). The lexicographic order function looks like a good candidate. If it is monotonic (it looks like so, but I'm not sure: \\(l(x, y) = x\\)) and doesn't obey \\(C^\times\\), then the conclusion is that the condition \\(C^\times\\) is supported only by a subset \\(F \subset M^\times\\).

So the question is whether any monotone will do to automatically transform \\((X \times X)\\) into preorder, or the condition brings something new to the table? Still figuring it out.

As we learned, the condition implies that the function is monotone: \\(C^\times \Rightarrow f \in M^\times\\), where \\(M^\times\\) is the set of monotone functions \\(X \times X \to X\\). The question is whether converse \\(f \in M^\times \Rightarrow C^\times\\) is also true, that is the function \\(f\\) being monotonic implies that it supports the condition \\(C^\times\\)?

The easiest way to show that converse is not true, is to find a monotone function \\(g \in M^\times\\) which doesn't support \\(C^\times\\). The lexicographic order function looks like a good candidate. If it is monotonic (it looks like so, but I'm not sure: \\(l(x, y) = x\\)) and doesn't obey \\(C^\times\\), then the conclusion is that the condition \\(C^\times\\) is supported only by a subset \\(F \subset M^\times\\).

So the question is whether any monotone will do to automatically transform \\((X \times X)\\) into preorder, or the condition brings something new to the table? Still figuring it out.