Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 121 / 3 / v / b / Solution

Assume that $\kappa$ is regular and uncountable in $M$, as required for this claim. If an antichain $A\in M$ had size $\kappa$, apply the delta-system lemma at a regular uncountable cardinal inside $M$ to the finite domains of its conditions. After thinning, their domains form a delta-system with finite root $R$.

There are fewer than $\kappa$ possible restrictions to $R$, because each coordinate $(\alpha ,n)\in R$ has fewer than $\kappa$ possible values. Since $\kappa$ is a regular cardinal, we can thin again to $\kappa$ conditions agreeing on $R$. Any two now have disjoint domains outside $R$ and agree on $R$, so their union is a common extension. This contradicts that $A$ is an antichain. ThereforeIf the word “regular” is allowed to include $\kappa =\omega$, the printed claim needs the additional hypothesis that $\kappa$ is uncountable; the finite-condition order can have an infinite antichain when $\kappa =\omega$.