Generalized delta systems for small supports

Well-order the family as $⟨x_{\xi }:\xi <\theta ⟩$. Its union has cardinality at most $\theta \cdot \kappa =\theta$, so transport its elements into $\theta$. Since $\theta$ is regular and there are only $\kappa <\theta$ possible order types below $\kappa$, thin to constant order type $\eta <\kappa$ and enumerate each remaining set increasingly as $x_{\xi }=\{x_{\xi }(i):i<\eta \}$.

Fewer than $\theta$ members of the thinned family can lie wholly below any fixed $\alpha <\theta$, because there are only $\mathrm{\mid }\alpha {\mathrm{\mid }}^{<\kappa }<\theta$ such sets. Its union is therefore unbounded in $\theta$, so some coordinate $i<\eta$ has unbounded values; let $i_0$ be least. For every $i<i_0$ the coordinate values are bounded, and regularity gives ${\alpha }_0=\sup \{x_{\xi }(i)+1:\xi <\theta ,i<i_0\}<\theta \mathrm{.}$ Recursively choose $x_{{\xi }_{\mu }}$ for $\mu <\theta$ so that $x_{{\xi }_{\mu }}(i_0)$ exceeds ${\alpha }_0$ and every element of the earlier chosen sets. The unboundedness of the $i_0$th values makes each choice possible. Consequently distinct chosen sets meet only below ${\alpha }_0$.

There are at most $\mathrm{\mid }{\alpha }_0{\mathrm{\mid }}^{<\kappa }<\theta$ possible intersections $x_{{\xi }_{\mu }}\cap {\alpha }_0$. Regularity therefore yields a set $J\subseteq \theta$ of size $\theta$ on which this intersection is one fixed $r$. Step 1.2 then gives $x_{{\xi }_{\mu }}\cap x_{{\xi }_{\nu }}=r$ for distinct $\mu ,\nu \in J$. AC was used to well-order, enumerate, choose recursively, and thin.

Suppose now that $\kappa$ is regular, $\rho =2^{<\kappa }$, and $\theta ={\rho }^{+}$. For every $\mu <\kappa$, a function $\mu \to \rho$ uses fewer than $\kappa$ members of the union $\rho ={\bigcup }_{\nu <\kappa }{}^{\nu }2$; regularity bounds all their lengths below one $\nu <\kappa$. Coding the function by a binary array of size below $\kappa$ shows ${\rho }^{\mu }\le 2^{<\kappa }=\rho$. Thus ${\rho }^{<\kappa }=\rho$, and for $\alpha <\theta$ one has $\mathrm{\mid }\alpha {\mathrm{\mid }}^{<\kappa }\le {\rho }^{<\kappa }=\rho <\theta$. The main clause applies.