Every ultrafilter on every set is principal in Blass's model

Let $D$ be the class of sets uniquely definable in $M[G]$ from $f$, finitely many members of $S\setminus \{f\}$, and finitely many ordinals, so that $N=\{x:\mathrm{TC}(\{x\})\subseteq D\}$. Since $S$ is definable from $f$, $D$ and $N$ are definable from $f$. Finite lists of definition parameters concatenate, exactly as in F9, so $D$ is closed under every fixed uniquely defined operation on finitely many members of $D$. The hereditary clause gives transitivity and all ordinals. Pairing, Union, Infinity, Extensionality and Foundation follow as in the ordinary HOD proof. For a fixed formula, Separation in $N$ is obtained by defining the required subset after relativizing quantifiers to the definable class $N$; internal Power Set is ${\mathcal{P}}^{M[G]}(x)\cap N$; and Replacement is the set of uniquely specified $N$-outputs. Substitution of the finite definitions of the parameters puts each resulting set in $D$, while transitivity puts all its descendants in $D$. Thus $N$ is a transitive ZF inner model of $M[G]$. No well-order of $S$ and no Choice in $N$ was used.

Suppose toward a contradiction that $U\in N$ is a free ultrafilter on $\omega$. By F3, a finite set in $U$ would put one of its singleton pieces in $U$, making $U$ principal; hence $U$ omits every finite set and contains every cofinite set. By F1, $U$ has a unique definition from $f$, ordinals, and finitely many reals $s_1,\dots ,s_t\in S\setminus \{f\}$. Each $s_i$ is a finite modification either of $a_{m_i}$ or of its complement. Choose $k=1+\max (\{m_1,\dots ,m_t\}\cup \{0\})$, so every coordinate occurring among the real parameters is below $k$. Let $A$ be whichever of $a_k$ and $\omega \setminus a_k$ belongs to $U$, as supplied by F3.

By F6, some finite $p\in G$ forces the unique defining formula for $U$ together with $A\in U$. Apply F5 with $n=k-1$: flip precisely the bits of coordinate $k$ above the finite domain of $p$. Since every $m_i<k$, this fixes $p$, every $s_i$, and all ordinals. It fixes $f$ because it interchanges the two finite-modification classes in $f(k)$ and fixes every other value. Its image $A^{\mathrm{\prime }}$ is equal modulo a finite set to $\omega \setminus A$. F8 therefore makes the same $p$ force $A^{\mathrm{\prime }}\in U$. In $M[G]$, both $A$ and $A^{\mathrm{\prime }}$ belong to $U$, so their finite intersection belongs to $U$, contradicting step 2.1. The involution treats the two possible choices of $A$ identically. Consequently every ultrafilter on $\omega$ in $N$ is principal.

Suppose now that some ordinal carries a free ultrafilter in $N$, and let $\kappa$ be the least such ordinal with witness $U$. Step 3.1 and transport along a bijection show that $\kappa$ is uncountable. Let $\gamma$ be the least ordinal for which there is a partition $⟨A_{\xi }:\xi <\gamma ⟩$ of $\kappa$ with every $A_{\xi }\mathrm{\notin }U$; it exists with $\gamma \le \kappa$ by the singleton partition. If $\gamma <\kappa$, the least-piece map $h:\kappa \to \gamma$ pushes $U$ to an ultrafilter $h_{\ast }U$ on $\gamma$. It cannot be principal, since $\{\xi \}\in h_{\ast }U$ would say $A_{\xi }\in U$. This contradicts the minimality of $\kappa$, so $\gamma =\kappa$. The same pushforward along a hypothetical bijection from $\kappa$ to a smaller ordinal shows that $\kappa$ is a cardinal.

The ultrafilter $U$ is uniform: if $B\in U$ had cardinality $\lambda <\kappa$, restricting $U$ to $B$ and transporting it along a bijection $B\to \lambda$ would give a free ultrafilter below $\kappa$. It is also $\kappa$-complete. Otherwise choose $\eta <\kappa$ and $B_{\xi }\in U$ for $\xi <\eta$ with $C={\bigcap }_{\xi <\eta }{B}_{\xi }\mathrm{\notin }U$. The sets consisting of points whose least failed membership test is $\xi$, together with $C$, partition $\kappa$ into $\eta +1<\kappa$ many $U$-small pieces: the $\xi$-piece is contained in $\kappa \setminus B_{\xi }$, and $C$ is small by assumption. This contradicts step 4.1. The empty intersection is $\kappa \in U$, so the argument includes $\eta =0$.

Fix a finite list of $S$-real and ordinal parameters uniquely defining $U$, and let $K\subseteq \omega$ be the finite set of their Cohen-coordinate indices. Put $V_0=M[G\upharpoonright (K\times \omega )]$ and factor the remaining forcing as $Q=\mathrm{Fn}((\omega \setminus K)\times \omega ,2)$. Since $M=L$, every member of the finite-real extension $V_0=L[⟨a_k:k\in K⟩]$ is hereditarily definable from those finitely many permitted reals and ordinals, so $V_0\subseteq N$. Let "$A\in \dot{U}$" abbreviate the forcing-language assertion that $A$ belongs to the unique object satisfying the fixed definition of $U$; this avoids choosing a noncanonical name. If $q,r\in Q$, flip the finitely many bits on which their common domains disagree; the image of $q$ is compatible with $r$. Such a flip fixes every ground name from $V_0$, fixes the defining real parameters, and fixes $f$ because finite changes preserve every $\delta$-class. Thus F8, persistence and density closure show that every assertion "$A\in \dot{U}$" with $A\in \mathcal{P}(\kappa {)}^{V_0}$ is decided by the top condition of $Q$.

In $V_0$ define $U_0=\{A\in \mathcal{P}(\kappa {)}^{V_0}:1_Q{⊩}_{Q}A\in \dot{U}\}\mathrm{.}$ The forcing relation is definable there, and F6 together with the homogeneity calculation in step 6.1 gives $U_0=U\cap \mathcal{P}(\kappa {)}^{V_0}$. Hence F3 transfers properness, complement decision, and nonprincipality to $U_0$. If $\eta <\kappa$ and a sequence $⟨A_{\xi }:\xi <\eta ⟩\in V_0$ consists of members of $U_0$, then the sequence belongs to $N$ because $V_0\subseteq N$; step 5.1 puts its intersection in $U$, and that intersection is computed in $V_0$. It therefore belongs to $U_0$. Thus $V_0$ regards $U_0$ as a nonprincipal $\kappa$-complete ultrafilter and regards $\kappa$ as measurable.

The forcing from $M$ to $V_0$ is trivial when $K=\mathrm{\varnothing }$ and otherwise countable. Any ground bijection witnessing that $\kappa$ was countable or was not a cardinal would remain a witness in $V_0$, so step 7.1 implies that $M$ already regards $\kappa$ as an uncountable cardinal; the forcing size is therefore below $\kappa$. F4 says that $M$ already has a measurable cardinal. Internally choose its least measurable cardinal $\lambda$ and use the auxiliary ultrapower construction in F4 to obtain a normal-measure embedding $j:M\to Q_0$ with critical point $\lambda$. Elementarity gives $Q_0\models V=L$. The transitive target contains every ordinal: for each ordinal $\alpha$, $j(\alpha )$ is an ordinal at least $\alpha$, so transitivity puts $\alpha$ in $Q_0$. F10 now gives $L^{Q_0}=L^M=M$, whence $Q_0=M$. But $\lambda$ is parameter-free definable as the least measurable cardinal, so elementarity and $Q_0=M$ give $j(\lambda )=\lambda$, contradicting that $\lambda$ is the critical point. This discharges the supposition in step 4.1: every ultrafilter on every ordinal in $N$ is principal.

Work henceforth inside the ZF model $N$. Let $W$ be the least class containing every singleton and closed under unions indexed by ordinals: equivalently, start with the empty set and all singletons and at each successor stage add every ${\bigcup }_{\alpha <\theta }{X}_{\alpha }$ whose pieces appeared earlier, taking unions at limit stages. The least construction stage is the $W$-rank. Thus every nonsingleton $X\in W$ has a presentation as a well-ordered union of sets of strictly smaller $W$-rank. This hierarchy is a definable class in $N$; it does not select presentations simultaneously.

Induction on $W$-rank proves three closure facts, with ranks no larger than those generated in the induction. If $B\subseteq {\bigcup }_{\alpha <\theta }{X}_{\alpha }$, then $B={\bigcup }_{\alpha <\theta }(B\cap X_{\alpha })$, and the induction hypothesis applies to each intersection; the empty and singleton bases are immediate. If $h:X\twoheadrightarrow Y$, then $Y={\bigcup }_{\alpha <\theta }h[X_{\alpha }]$, giving closure under surjective images by the same induction. A double induction gives finite products: distribute $X\times Y$ over a well-ordered-union presentation in either coordinate, with singleton and empty products as bases. Consequently finite sequences from a $W$-set, graphs and relations cut out as subsets of finite products, and well-ordered unions of these objects also lie in $W$.

For each fixed $n$, the map sending a finite subset $z\subseteq \omega$ to $a_n\mathrm{△}z$ enumerates $\delta (a_n)$; the analogous map enumerates $\delta (\omega \setminus a_n)$. These maps exist in $N$ using the single permitted parameter $a_n$ and the canonical well-order of the finite subsets of $\omega$. Hence each class is well-orderable and lies in $W$, without choosing representatives for all classes at once. The pair $C_n=\delta (a_n)\cup \delta (\omega \setminus a_n)$ lies in $W$, and the sequence $n\mapsto C_n$ is definable from $f$. Therefore $S=\{f\}\cup {\bigcup }_{n<\omega }{C}_n\in W\mathrm{.}$

We have $W\subseteq N$ because $N$ is transitive, contains all its singletons, and is internally closed under well-ordered unions. Conversely fix $x\in N$. For each $y\in x$, let ${\theta }_y$ be the least ambient hierarchy bound at which some formula, finite ordinal tuple, and finite tuple from $S$ uniquely define $y$ from $f$; the set-level satisfaction coding used in F9 makes this a set-theoretic predicate. Replacement bounds the ${\theta }_y$ by one ordinal $\theta$. Let $D$ be the set of all bounded definition codes whose unique output belongs to $x$. Substituting the fixed finite-parameter definition of $x$ shows that $D$ and its evaluation map are in $N$; no code was chosen separately for each $y$. The code space is a subset of a finite product and a well-ordered union of ${\theta }^{<\omega }$, $\omega$, and $S^{<\omega }$, so steps 10.1--11.1 put $D$ in $W$. Evaluation maps $D$ onto $x$, and closure under surjective images puts $x$ in $W$. Thus $N=W$.

Induct on the $W$-rank of a carrier $X$. There is no proper ultrafilter on $\mathrm{\varnothing }$, and every ultrafilter on a singleton is principal. Otherwise use step 9.1 to write $X={\bigcup }_{\alpha <\theta }{X}_{\alpha }$ with lower-rank pieces and refine it to the disjoint partition $Y_{\alpha }=X_{\alpha }\setminus {\bigcup }_{\beta <\alpha }{X}_{\beta }$; step 10.1 keeps every $Y_{\alpha }$ at lower rank. For an ultrafilter $U$ on $X$, the least-piece map $h:X\to \theta$ pushes $U$ to an ultrafilter on the ordinal $\theta$. By step 8.1 it is principal, say at ${\alpha }_0$, so $Y_{{\alpha }_0}=h^{-1}(\{{\alpha }_0\})\in U$ and is nonempty. The restriction of $U$ to $Y_{{\alpha }_0}$ is an ultrafilter there and is principal at some $y$ by induction. For every $B\subseteq X$, upward closure and intersection give $B\in U⟺B\cap Y_{{\alpha }_0}\in U⟺y\in B\mathrm{.}$ Hence $U$ is principal at $y$. Step 12.1 says every carrier in $N$ has a $W$-rank, so this proves the theorem for every set in $N$.