Small forcing does not create measurable cardinals

The trivial-forcing case is immediate, so suppose $P$ is nontrivial and replace it by an isomorphic forcing on an ordinal of size $\mu =\mathrm{\mid }P{\mathrm{\mid }}^V$. A measure on $\kappa$ is uniform: if $A$ in the measure had size $\lambda <\kappa$, intersecting the complements of its $\lambda$ singletons would both retain $A$ and make the intersection empty. Uniformity and $\kappa$-completeness make $\kappa$ regular, since the bounded pieces of a cofinal partition of length below $\kappa$ would all be measure-small. They also make $\kappa$ a strong limit: if $\kappa \le 2^{\lambda }$ for $\lambda <\kappa$, choose $\kappa$ distinct binary subsets of $\lambda$; for every coordinate take the bit occurring on a measure-one set and intersect these fewer than $\kappa$ sets. The intersection has at most one member, contradicting uniformity. Thus $\kappa$ is strongly inaccessible in $V[G]$. Forcing of size $\mu$ is ${\mu }^{+}$-cc and preserves cardinals at and above ${\mu }^{+}$, so $\kappa$ is also a ground cardinal. Choose a regular ground cardinal $\delta$ with $\mu <\delta <{\delta }^{+}<\kappa \mathrm{.}$

We give the ultrapower facts needed later. From a nonprincipal $\kappa$-complete ultrafilter $U$ on $\kappa$, form the ultrapower using functions $\kappa \to V[G]$. Łoś's induction uses F1 and the witness choices supplied by F3. It is well-founded: an external descending omega-sequence would, by countable completeness, give one coordinate carrying an infinite descending sequence of ordinals. After transitive collapse, the constant-function map $j:V[G]\to \bar{M}$ is elementary, fixes every ordinal below $\kappa$, and moves $\kappa$, so its critical point is $\kappa$. The derived measure $W=\{A\subseteq \kappa :\kappa \in j(A)\}$ is normal: for a regressive $f$ on a $W$-large set, $j(f)(\kappa )<\kappa$ is fixed by $j$, and its fibre is $W$-large. Take the ultrapower by $W$ and again call its collapsed map $j$. Its target is closed under $\kappa$-sequences from $V[G]$: given $x_{\alpha }=[f_{\alpha }{]}_W$ for $\alpha <\kappa$, use F3 to choose the representatives and define $F(\xi )=⟨f_{\alpha }(\xi ):\alpha <\xi ⟩$. Normality identifies the seed $[\mathrm{i}\mathrm{d}{]}_W$ with $\kappa$, and $j(F)(\kappa )(\alpha )=x_{\alpha }$ for every $\alpha <\kappa$, so the entire sequence belongs to the target. Conversely, for any definable elementary $j$ into a transitive class with critical point $\kappa$, the same seed formula defines a set-sized nonprincipal $\kappa$-complete ultrafilter: elementarity gives complement decision and intersections, and $j$ fixes all singleton indices below $\kappa$.

Thus forcing by $P$ is forcing with a gap at $\delta$: the initial forcing has size below $\delta$ and the tail forcing is trivial, hence $\le \delta$-strategically closed.

By step 1.2, in $V[G]$ take a normal-measure ultrapower $j:V[G]⟶\bar{M}$ with critical point $\kappa$. The transitive target is closed under $\kappa$-sequences of the extension and hence under $\delta$-sequences. As in the general setup of the Gap Forcing Theorem, define the ground part $M={\bigcup }_{\alpha \in \mathrm{O}\mathrm{r}\mathrm{d}}j(V_{\alpha }^V)$, taking the transitive collapse implicit in this notation. Then $j\upharpoonright V:V\to M$, $j(G)$ is $M$-generic, and $\bar{M}=M[j(G)]$. By the ordinal presentation chosen in step 1.1, $P\in V_{\kappa }$, so $j(P)=P$; the image-filter calculation gives $j(G)=G$, and consequently $\bar{M}=M[G]$. The critical-point calculation also gives agreement below $\kappa$, $V_{\kappa }=M_{\kappa }$, in the two ground parts. The remaining task is to prove that $M$ and the restricted map actually belong to the ground model, not merely to the forcing extension.

First record the fresh-sequence obstruction specialized to small forcing. If $\mathrm{cf}(\theta )>\mu$, $P$ adds no sequence $s:\theta \to \mathrm{O}\mathrm{r}\mathrm{d}$ which is new while every proper initial segment is in $V$. Indeed, for each $\alpha <\theta$ the truth lemma gives a condition $p_{\alpha }\in G$ and a ground sequence $s_{\alpha }$ such that $p_{\alpha }⊩\dot{s}\upharpoonright \alpha ={\check{s}}_{\alpha }$. One condition $p\in G$ occurs for an unbounded set of $\alpha$, because there are at most $\mu$ conditions and $\mathrm{cf}(\theta )>\mu$. Persistence then makes $p$ decide all of $\dot{s}$ as the union of those compatible ground initial segments, contrary to newness. The same argument works over $M$ for the forcing $j(P)=P$.

We next prove the common-cover claim used by the restriction. If $\sigma$ is a set of ordinals of extension-cardinality $\delta$, there is a set $\tau \in M\cap V$ of cardinality $\delta$ with $\sigma \subseteq \tau$. First, a $\delta$-enumeration of $\sigma$ is a $\delta$-sequence of ordinals, so the closure from step 3.1 puts it, and hence $\sigma$, in $M[G]$. A $P$-name for such an enumeration has at most $\delta \mu =\delta$ possible ordinal values, so $\sigma$ has a $V$-cover of size $\delta$; the same name calculation in $M$ gives an $M$-cover. Alternate these two operations for $\delta$ stages, taking increasing covers, and let $\tau$ be their union. The resulting sequence belongs to $M[G]$ by its $\delta$-closure. On the cofinally many stages whose values lie in $V$, a single condition of $G$ decides unboundedly many values, because $\mathrm{\mid }P\mathrm{\mid }<\delta$ and $\delta$ is regular; monotonicity of the sequence makes that condition decide the union, so $\tau \in V$. Repeating this argument with an $M$-name at the cofinally many $M$-stages gives $\tau \in M$.

It follows that $M$ and $V$ have the same $\delta$-sequences of ordinals. For a size-$\delta$ set of ordinals $\sigma$ in either class, take the common cover $\tau$ from step 4.2 and enumerate it increasingly in both classes as $⟨{\beta }_{\xi }:\xi <\gamma ⟩$, where $\gamma <{\delta }^{+}<\kappa$. The index set $A=\{\xi <\gamma :{\beta }_{\xi }\in \sigma \}$ lies below $\kappa$. The agreement $V_{\kappa }=M_{\kappa }$ from step 3.1 puts $A$, and hence $\sigma$, in both classes. Shorter sequences are padded to length $\delta$.

We now show $M\subseteq V$. It suffices, by coding, to prove this for sets of ordinals. Induct on $\theta$ for $A\subseteq \theta$ in $M$, assuming every proper initial segment is in $V$. If $\mathrm{cf}(\theta )\ge \delta$, then a new $A$ would be a fresh $\theta$-sequence, contradicting step 4.1. If $\mathrm{cf}(\theta )<\delta$, write $A={\dot{A}}_G$ and choose a sufficiently large $V_{\zeta }$ with an elementary $X\prec V_{\zeta }$ of size $\delta$ containing $P$, every element of $P$, and $\dot{A}$. The set $X\cap \mathrm{O}\mathrm{r}\mathrm{d}$ belongs to $M$ by step 5.1. Hence $a=A\cap X$ is in $M$, and step 5.1 puts this size-at-most-$\delta$ set of ordinals in $V$. Some $p\in G$ forces $X\cap \dot{A}=\check{a}$. Thus $X$ satisfies that $p$ decides every membership question for $\dot{A}$ whose index lies in $X$; elementarity makes the same statement true in $V_{\zeta }$. Therefore $p$ decides all of $\dot{A}$, so $A\in V$. The usual membership-rank coding then yields $M\subseteq V$.

The identical fresh-sequence induction, now between $M$ and $M[G]$, shows $M=V\cap M[G]\mathrm{.}$ For a set of ordinals common to $V$ and $M[G]$, use step 4.1 at cofinality at least $\delta$ and step 5.1 at smaller cofinality; an arbitrary set is reduced to its index set in an $M$-enumeration of an ambient $M$-set. This is the exact target-identification needed below.

The ultrapower embedding is amenable to $V[G]$. We prove that $j\upharpoonright V$ is amenable to $V$. It is enough to show $j\mathrm{‘}\mathrm{‘}\theta \in V$ for every ordinal $\theta$. Induct on $\theta$. At cofinality at least $\delta$, a new image sequence would violate step 4.1. At smaller cofinality choose $X\prec V_{\zeta }$ as in step 6.1. The set $a=(j\mathrm{‘}\mathrm{‘}\theta )\cap X$ has size at most $\delta$; because $a$ is a small subset of $j\mathrm{‘}\mathrm{‘}\theta$, write $a=j\mathrm{‘}\mathrm{‘}b=j(b)$ for some $b\subseteq \theta$ of size at most $\delta$. Use step 4.2 to cover $b$ by a size-$\delta$ set $c\in M\cap V$ and replace $c$ by $c\cap \theta$. Then $a\subseteq (j\mathrm{‘}\mathrm{‘}c)\cap X\subseteq (j\mathrm{‘}\mathrm{‘}\theta )\cap X=a$, while $j\mathrm{‘}\mathrm{‘}c=j(c)\in M\subseteq V$, so $a\in V$. A condition in $G$ decides this trace, and elementarity of $X$ makes it decide the entire image sequence. Thus $j\mathrm{‘}\mathrm{‘}\theta \in V$. Replacement converts these image sequences to every set restriction $j\upharpoonright A\in V$.

In the ground model define $U_0=\{A\subseteq \kappa :A\in V\text{ and }\kappa \in j(A)\}$. Step 7.2 makes this a ground set. As in step 1.2, elementarity gives complement decision, finite-intersection closure and upward closure; no singleton belongs to $U_0$ because $j$ fixes all ordinals below $\kappa$. If $\eta <\kappa$ and $A_{\xi }\in U_0$ for every $\xi <\eta$, then $j(\eta )=\eta$ and $\kappa$ belongs to every $j(A_{\xi })$, hence to $j({\bigcap }_{\xi <\eta }{A}_{\xi })$. Thus $U_0$ is a nonprincipal $\kappa$-complete ultrafilter on $\kappa$ in $V$. By the local definition in the Statement, $\kappa$ was measurable in $V$, as required.