Alex wrote:

> MD 2. By the method in Lecture 6, r(y,z) = least upper bound of X = {x:Δ(x)≤(y,z)}

> * Since X is the set of elements of A less than min(y,z), r(y,z) is min(y,z).

This looks very good - except for one thing. Your calculation was right, but you jumped to a conclusion at the end.

In a totally ordered set either \\(y \\le z\\) or \\(y \le z\\) or both (in which case \\(y = z\\), so the **minimum** min(y,z) exists: it's the smaller one of \\(y\\) and \\(z\\) (or if they're equal, it's both).

But Matthew Doty's puzzle is extremely interesting, perhaps even more interesting, when our poset is _not_ totally ordered. In this case \\(\textrm{min}(y,z)\\) is no longer the best answer to the puzzle, because the minimum may not exist, but the answer may still exist.

(For example, consider a poset \\( P X \\) of all subsets of \\(X\\). This is not totally ordered, so it's easy to have two subsets \\(S , T \subseteq X \\), neither of which is smaller than the other.)

Similarly for this:

> MD 3. By duality, l(y,z) = max(y,z).