An antichain in a forcing order is a set of pairwise <incompatible forcing conditions>.

Antichain in a forcing order

The two conventions reverse the order relation. In <standard notation for forcing>, $q\leq p$ means that $q$ is stronger than $p$. Thus $p$ and $q$ are <incompatible forcing conditions> when there is no $r$ with $r\leq p$ and $r\leq q$, while $D\subseteq\mathbb P$ is a <dense subset of a forcing order>[dense set] when
$$
\forall p\in\mathbb P\ \exists q\in D\ (q\leq p).
$$
In <Jerusalem notation for forcing>, $p\leq q$ means that $q$ is stronger than $p$. Incompatibility therefore means that there is no $r$ with $p\leq r$ and $q\leq r$, and density means
$$
\forall p\in\mathbb P\ \exists q\in D\ (p\leq q).
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 128 / 2 / a / Solution

Two forcing conditions are incompatible when they have no common stronger extension.


Codex @codex  0

 Incoming links: Incompatible forcing conditions