Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 121 / 4

A forcing atom is a condition $p\in \mathbb{P}$ such that any two strengthenings of $p$ are compatible forcing conditions:An atomless forcing order has no such condition; equivalently,The meaning of $q\perp r$ is that there is no common stronger condition. For an arbitrary partial order, an atom need not be a minimal element: the definition concerns compatibility below it. This distinction prevents a minimal-element definition from misclassifying an order with descending but mutually compatible conditions.

The two-argument version of Fn forcing consists of finite partial functions:Thus a stronger condition specifies more values. The empty partial function is the greatest condition. Two conditions are compatible forcing conditions exactly when they agree on their common domain; if they do, their set union is a common strengthening. This is $\mathrm{Fn}(I,J,\omega )$ in the three-argument convention for Fn forcing. If $I$ or $J$ is empty, only the empty condition exists. For infinite $I$ and at least two elements in $J$, assigning two different values at a fresh coordinate proves that this is an atomless forcing order.

Suppose, towards a contradiction, that $G\in M$. Then $D=\mathbb{P}\setminus G$ is a ground-model set by axiom schema of separation. It is dense. If $p\notin G$, it already lies in $D$. If $p\in G$, atomless forcing order structure gives incompatible forcing conditions $q,r\le p$. Both cannot belong to the directed filter $G$, so at least one is a strengthening of $p$ in $D$.

All compatibility and extension quantifiers range over the same ground-model set $\mathbb{P}$, so this density argument is also valid internally in the transitive model $M$. Genericity requires $G\cap D\ne \varnothing$, contradicting the definition of $D$. Therefore a generic filter for an atomless order is new:The assumption is essential: a forcing atom determines a ground-model generic filter. For an atom $p$, the conditions compatible with $p$ form a generic filter $G_p\in M$: two such conditions have strengthenings below $p$, which have a common strengthening by the atom property; every dense set has a member below $p$. In a general partial order, $G_p$ need not be the principal filter above $p$.

Put $\kappa ={\aleph }_2^M$ and let $F=\bigcup G$. Conditions from Fn forcing agree on common coordinates because $G$ is directed, so $F$ is a function. For each $(\xi ,n)\in \kappa \times \omega$, the ground-model setis dense: add a value at that coordinate if necessary. The generic filter meets all these sets, making $F:\kappa \times \omega \to 2$ total.

For distinct $\xi ,\eta <\kappa$, the ground-model setis dense. Choose $n$ fresh for both rows of a finite condition and assign $0$ and $1$ there. Genericity therefore makes the rows pairwise different. The generic coordinate reals for finite-function forcing are $c_{\xi }=\{n\in \omega :F(\xi ,n)=1\}$, and in the generic extensionThe family is a set in $M[G]$, for example by collecting the ground-model names ${\dot{c}}_{\xi }=\{(\check{n},p):p(\xi ,n)=1\}$ and evaluating the corresponding name for their indexed family. This argument only needs $\kappa$ as the specified ground-model ordinal; it does not silently assume a cardinal preservation by chain-condition forcing result.

The Rasiowa–Sikorski lemma constructs a generic filter $G_i$ over each $M_i$. Every resulting generic extension remains a countable transitive model of ZFC: there are only countably many ground-model names externally. The order $\mathbb{P}$ has the same elements and ordering at every stage, and atomless forcing order structure is absolute because all its quantifiers are bounded to $\mathbb{P}$. Hence the generic filter for an atomless order is new result gives $G_i\notin M_i$ for every $i$.

The increasing union $N={\bigcup }_{i<\omega }{M}_i$ is transitive and contains $\mathbb{P}$. If it satisfied Axiom of power set for $\mathbb{P}$, there would be a set $A\in N$ withChoose $i$ with $A\in M_i$. The next-stage generic filter $G_i$ belongs to $M_{i+1}\subseteq N$ and is an actual subset of $\mathbb{P}$. This subset assertion is absolute for the transitive set $N$, so $G_i\in A$. Transitivity of $M_i$ and $A\in M_i$ then give $G_i\in M_i$, a contradiction. Thus the power-set failure in an increasing union of generic extensions occurs already at the fixed ground-model order:The stages form an increasing chain, not an elementary chain, so the elementary chain theorem does not assert ZFC for their union.