 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/iii/paper-121/3/iv/a/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

G_{0}

and

G_{1}

H

G_{0} = {p : \exists q ((p, q) \in H)}, G_{1} = {q : \exists p ((p, q) \in H)} .

These are nonempty upward-closed, directed filters in an ordered set. Clearly

H \subseteq G_{0} \times G_{1}

. If

p \in G_{0}

and

q \in G_{1}

, choose

(p, q^{'}) \in H

and

(p^{'}, q) \in H

. Directedness supplies

(r, s) \in H

below both, so

(r, s) \leq (p, q)

. Upward closure of

H

gives

(p, q) \in H

H = G_{0} \times G_{1} .

No greatest conditions in the factors are required for this argument.

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