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

The axiom of constructibility asserts every set belongs to the constructible universe, or $V=L$. Define $L_0=\varnothing$, $L_{\alpha +1}=\mathrm{Def}(L_{\alpha })$, and $L_{\eta }={\bigcup }_{\alpha <\eta }{L}_{\alpha }$ at limit stages. Here $\mathrm{Def}(X)$ consists of subsets of $X$ first-order definable over $(X,\in )$ with finitely many parameters from $X$. Then $L={\bigcup }_{\alpha \in \mathrm{Ord}}{L}_{\alpha }$. The assertion is $\forall x\exists \alpha (x\in L_{\alpha })$, rather than a statement that every set is parameter-free definable.

The Bukovský-Hechler theorem states: if $\kappa$ is singular and there are a cardinal $\rho <\kappa$ and a cardinal $\tau$ such that $2^{\mu }=\tau$ for every cardinal $\mu$ with $\rho \le \mu <\kappa$, thenThus an eventual plateau of the power-set function below a singular cardinal continues at that cardinal. One can see the mechanism by decomposing $\kappa$ into $\mathrm{cf}(\kappa )$ bounded pieces: $2^{\kappa }=(2^{<\kappa }{)}^{\mathrm{cf}(\kappa )}$. On the plateau choose $\mu \ge \mathrm{cf}(\kappa )$, so ${\tau }^{\mathrm{cf}(\kappa )}=(2^{\mu }{)}^{\mathrm{cf}(\kappa )}=2^{\mu }=\tau$.

The singular cardinals hypothesis is the assertion that for every infinite singular cardinal $\kappa$,In particular, for a singular strong-limit cardinal it gives $2^{\kappa }={\kappa }^{+}$. This is a restriction on singular-cardinal exponentiation; it does not assert the full Generalized continuum hypothesis at all cardinals.

Put $\theta =\mathrm{cf}(\delta )$. There is a strictly increasing cofinal map $a:\theta \to \delta$. If $\mathrm{cf}(\theta )=\nu <\theta$, choose a cofinal map $b:\nu \to \theta$. Then $a\circ b$ is cofinal in $\delta$, contradicting the definition of $\theta$ as its least cofinal order type. Since always $\mathrm{cf}(\theta )\le \theta$, we obtainFor a nonzero limit ordinal this is an infinite regular cardinal. If the convention includes zero among limit ordinals, its cofinality is zero and the identity is immediate there.

First the increasing sequence is cofinal in $\kappa$. Write $\mu ={\sup }_{\xi <\delta }{\kappa }_{\xi }$. Strict increase gives $|\delta |\le \mu$ when the sequence is infinite, so its cardinal sum is $\max (|\delta |,\mu )=\mu$. The given sum therefore implies $\mu =\kappa$ and $|\delta |\le \kappa$.

For any function $f:\lambda \to \kappa$, choose an index ${\xi }_t$ with $f(t)<{\kappa }_{{\xi }_t}$ for every $t<\lambda$. There are at most $\lambda <\mathrm{cf}(\delta )$ such indices, so they are bounded below $\delta$. Enlarging the bound if needed gives an index $\xi$ with $\mathrm{ran}(f)\subseteq {\kappa }_{\xi }$. ThusConversely each summand is at most ${\kappa }^{\lambda }$, and $|\delta |\le \kappa \le {\kappa }^{\lambda }$ because $\lambda \ne 0$. Infinite cardinal multiplication consequently givesCombining the inequalities proves ${\kappa }^{\lambda }={\sum }_{\xi <\delta }{\kappa }_{\xi }^{\lambda }$.

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.