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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/iii/paper-121/4/i/solution

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Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 121 / 4 / i / Solution by

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

p \in P

such that any two strengthenings of

p

are compatible forcing conditions:

p is an atom ⟺ \forall q, r \leq p \exists s \in P (s \leq q, r) .

(1)

An atomless forcing order has no such condition; equivalently,

\forall p \in P \exists q, r \leq p (q ⊥ r) .

(2)

The meaning of

q ⊥ r

is that there is no common stronger condition. For an arbitrary partial order, an atom need not be a minimal element: the definition concerns compatibility below it. This distinction prevents a minimal-element definition from misclassifying an order with descending but mutually compatible conditions.

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