Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 19 / 6 / ii

Choose a name $\dot{f}\in M$ for $f$ and a condition $p\in G$ forcing that it is a function $\alpha \to \beta$. The ordinals $\alpha ,\beta$ belong to $M$, by forcing preserves ordinals. For each $\zeta <\alpha$, choose in $M$ a maximal forcing antichain $A_{\zeta }$ in the cone above $p$, every member deciding $\dot{f}(\zeta )$ as an ordinal below $\beta$. Conditions deciding an ordinal-valued name are dense, by the forcing theorem.

Let $y(\zeta )$ be the set of values decided by members of $A_{\zeta }$. The chain condition gives $|y(\zeta ){|}^M<\kappa$, and Choice and Replacement in $M$ assemble all these sets into a function $y\in M$ with domain $\alpha$. Since $p\in G$, genericity ensures that $G$ meets each deciding forcing antichain above $p$; equivalently enlarge it to a global maximal forcing antichain by conditions incompatible with $p$. Its chosen value is the actual $f(\zeta )$. ThereforeThis is the possible-values lemma for chain-condition forcing. All size bounds in its construction are internal to the ground model.

Suppose a ground cardinal $\lambda \ge \kappa$ became smaller in the extension. There would be a surjection $f:\alpha \to \lambda$ for some ordinal $\alpha <\lambda$. A finite domain cannot map onto an infinite ordinal, so use part (a) to cover $f$ by a ground function $y$ of small value sets. Its range is contained in $U={\bigcup }_{\zeta <\alpha }y(\zeta )$.

If $\lambda >\kappa$, the ground cardinality of $U$ is at most $|\alpha {|}^M\cdot \kappa <\lambda$. If $\lambda =\kappa$, regularity bounds a union of fewer than $\kappa$ sets of size less than $\kappa$ by a cardinal less than $\kappa$. In either case $U$ is a proper ground subset of $\lambda$, and the same missing ordinal remains missing in the extension, contradicting surjectivity.

Hence all ground cardinalities at least $\kappa$ are preserved. Ground bijections also remain bijections, so old noncardinals cannot turn into cardinals. The assertion does not prevent collapse of smaller cardinals, and consequently does not by itself preserve the aleph index of $\kappa$.