Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 19 / 2 / iv

For the intended infinite regular cardinal $\kappa$, set $S_0(\delta )=0$. Suppose $0<\zeta <{\kappa }^{+}$ and the functions for its predecessors have been constructed. Since $|\zeta |\le \kappa$, choose a surjection $e_{\zeta }:\kappa \to \zeta$ and defineThe supremum is below $\kappa$: it uses fewer than $\kappa$ ordinals below $\kappa$, and $\kappa$ is regular. This defines a function $\kappa \to \kappa$ at each stage of the recursion.

For any $\alpha <\zeta$, choose $\eta$ with $e_{\zeta }(\eta )=\alpha$. Whenever $\delta >\eta$, the displayed supremum includes $S_{\alpha }(\delta )+1$, so $S_{\alpha }(\delta )<S_{\zeta }(\delta )$. Therefore the recursion yields a long chain under eventual domination of length ${\kappa }^{+}$.

Take the graphs $G_{\zeta }=\{(\delta ,S_{\zeta }(\delta )):\delta <\kappa \}$. Each has size $\kappa$, and any two meet in fewer than $\kappa$ points, because their functions are eventually strictly ordered. Fix a bijection $b:\kappa \times \kappa \to \kappa$ and let $A_{\zeta }=b[G_{\zeta }]\in [\kappa {]}^{\kappa }$.

Given $I\in [{\kappa }^{+}{]}^{\kappa }$, enumerate it as $\{{\zeta }_{\eta }:\eta <\kappa \}$ and trim the sets in this order:The deleted part is a union of fewer than $\kappa$ pairwise intersections, each of size less than $\kappa$. Regularity ensures its size is less than $\kappa$. If $\nu <\eta$, then $A_{{\zeta }_{\eta }}^{\ast }$ is disjoint even from the full earlier $A_{{\zeta }_{\nu }}$, hence from its trimmed version. Thus all the retained sets are pairwise disjoint and each lost fewer than $\kappa$ elements. This provesThe argument is the essential disjointness of small subfamilies of a regular-cardinal almost disjoint family.