Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 121 (Topics in Set Theory)

For an infinite cardinal number $\kappa$, the gimel function isThe singular cardinals hypothesis asserts that for every infinite singular cardinal $\kappa$,Here ${\kappa }^{+}$ is the successor cardinal. Equivalently, for every infinite singular cardinal,To see why the second formulation adds nothing in the other case, put $\theta =\mathrm{cf}(\kappa )$. If $\kappa \le 2^{\theta }$, then $2^{\theta }\le {\kappa }^{\theta }\le (2^{\theta }{)}^{\theta }=2^{\theta }$ by infinite cardinal arithmetic. Equality $\kappa =2^{\theta }$ is impossible here: the König theorem for cardinal numbers gives $\mathrm{cf}(2^{\theta })>\theta$, whereas $\mathrm{cf}(\kappa )=\theta$. Thus in this case $2^{\theta }\ge {\kappa }^{+}$, as required by the maximum formula. For a strong limit cardinal that is singular, the hypothesis also yields $2^{\kappa }={\kappa }^{+}$: restrictions of a subset of $\kappa$ to a cofinal sequence of smaller cardinals give $2^{\kappa }\le {\kappa }^{\theta }$, while the reverse inequality is immediate. The quantified implication above is the precise general statement.

Write $\mathfrak{c}=2^{{\aleph }_0}$. Its cofinality is ${\aleph }_1$, so choose a strictly increasing sequence of infinite cardinal numbers $({\kappa }_{\xi }{)}_{\xi <{\omega }_1}$ below $\mathfrak{c}$ with supremum $\mathfrak{c}$. Such a sequence is obtained by refining a cofinal sequence and choosing larger cardinals recursively; $\mathfrak{c}$ is a singular cardinal, hence a limit cardinal.

By the axiom of choice, fix a bijection $b:\mathfrak{c}\to \mathbb{R}$. SetEvery $A_{\xi }$ is infinite, $|A_{\xi }|={\kappa }_{\xi }<\mathfrak{c}$, and distinct indices give distinct cardinalities. The cofinality of the sequence ensuresThus $|\mathcal{A}|={\aleph }_1$. Pairwise different sizes means different sizes for distinct members; without that qualification the parenthetical condition would contradict the case $A=B$.

The least index is $\boxed{\alpha ={\omega }_1}$, withIndeed, for every nonzero limit ordinal $\beta$, continuity of the aleph number enumeration and the cofinality of an increasing ordinal supremum giveIf $\beta <{\omega }_1$ is a nonzero limit ordinal, this cofinality is $\omega$, so ${\aleph }_{\beta }$ is a singular cardinal. The other infinite cardinal numbers below ${\aleph }_{{\omega }_1}$ are successor-indexed or ${\aleph }_0$, and are regular cardinals. For a successor cardinal ${\mu }^{+}$, a cofinal sequence of length at most $\mu$ would express ${\mu }^{+}$ as a union of at most $\mu$ sets of size at most $\mu$, contradicting infinite cardinal arithmetic.

There are ${\aleph }_1$ nonzero countable limit ordinals, and each countable initial segment contains only countably many. In increasing order they therefore have order type ${\omega }_1$. These are precisely the indices of the uncountable singular cardinals below ${\aleph }_{{\omega }_1}$. Hence every ${\sigma }_{\alpha }$ with $\alpha <{\omega }_1$ has countable cofinality, and the next one is ${\aleph }_{{\omega }_1}$, whose cofinality is ${\omega }_1<{\aleph }_{{\omega }_1}$.

Yes. A cardinal fixed point of the singular cardinal enumeration is obtained by countable iteration. Start with ${\kappa }_0={\aleph }_0$ and putThe singular cardinal enumeration is strictly increasing and satisfies ${\sigma }_{\alpha }\ge \alpha$ for every ordinal $\alpha$; the latter follows by transfinite induction for any strictly increasing ordinal-valued enumeration. If equality occurs at some ${\kappa }_n$, that cardinal number is already a witness. Otherwise the sequence is strictly increasing. Put $\kappa ={\sup }_{n<\omega }{\kappa }_n$. This is an uncountable singular cardinal of cofinality $\omega$.

For every $\alpha <\kappa$, some $n$ has $\alpha <{\kappa }_n$, and thereforeAlso ${\sup }_n{\sigma }_{{\kappa }_n}=\kappa$, so ${\sup }_{\alpha <\kappa }{\sigma }_{\alpha }=\kappa$. At a limit ordinal index, if the supremum of all preceding enumerated cardinals is itself singular, it is exactly the next member: every smaller singular cardinal already has a preceding index. ThusThis argument uses continuity only at a singular supremum; the enumeration need not be continuous at a weakly inaccessible cardinal.

It suffices to obtain a model of ZFC without weakly inaccessible cardinals. Starting with any model of ZFC, pass to its constructible universe, which satisfies ZFC and the Generalized continuum hypothesis. If it has no inaccessible cardinal, use that model. Otherwise pass to its rank segment at its least inaccessible cardinal $\delta$. This segment satisfies ZFC, retains the Generalized continuum hypothesis, and has no inaccessible cardinals. Under the Generalized continuum hypothesis, every weakly inaccessible cardinal is strongly inaccessible: if $\lambda <\kappa$ and $\kappa$ is a limit cardinal, then $2^{\lambda }={\lambda }^{+}<\kappa$. Thus in either case the resulting model $N$ has no weakly inaccessible cardinals.

Work inside $N$. If $\alpha$ is a nonzero limit ordinal, let $\lambda ={\sup }_{\beta <\alpha }{\sigma }_{\beta }$. It is an uncountable limit cardinal. If it were regular, it would be a weakly inaccessible cardinal, which is impossible in $N$. It is therefore singular, and the singular cardinal enumeration is continuous at this index:The cofinality of an increasing ordinal supremum now givesfor every nonzero limit ordinal $\alpha$ in $N$. Hence $N$ satisfies the negation of the proposed existential assertion. By the soundness theorem for first-order logic, consistency of ZFC prevents ZFC from proving that assertion. The model construction is a relative-consistency argument; it does not assume that consistency alone supplies a countable transitive model. Here, as usual, a limit ordinal excludes zero.

The Freiling axiom of symmetry isThe values of $f$ are countable subsets of the real numbers, including finite ones. One may require $x\ne y$ without changing the axiom: apply the displayed version to $x\mapsto f(x)\cup \{x\}$.

The Generalized continuum hypothesis statesHere ${\kappa }^{+}$ is the successor cardinal; equivalently $2^{{\aleph }_{\alpha }}={\aleph }_{\alpha +1}$ for every ordinal $\alpha$.

The Freiling theorem givesFirst assume the Continuum hypothesis and fix a well-order $(r_{\alpha }{)}_{\alpha <{\omega }_1}$ of the real numbers. Define $f(r_{\alpha })=\{r_{\beta }:\beta \le \alpha \}$. Each value is countable. For any pair, one index is at most the other, so at least one of $x\in f(y)$ or $y\in f(x)$ holds. The same holds for $x=y$, since $x\in f(x)$. This violates the Freiling axiom of symmetry.

Conversely, assume $|\mathbb{R}|>{\aleph }_1$ and let $f$ assign a countable set to each real. Choose $X\subseteq \mathbb{R}$ of cardinality ${\aleph }_1$. By infinite cardinal arithmetic,Choose $y$ outside this union. Since $f(y)$ is countable and $X$ is uncountable, choose $x\in X\setminus f(y)$. Then $y\notin f(x)$ and $x\notin f(y)$, with $x\ne y$. This proves the Freiling axiom of symmetry and completes both implications.

The construction gives a Lusin set $\boxed{A=\{x_{\alpha }:\alpha <{\omega }_1\}}$ with all $x_{\alpha }$ distinct.

Every meagre set is contained in a meagre $F_{\sigma }$ set: replace each nowhere dense set in a countable covering by its closure. There are at most $2^{{\aleph }_0}$ such covers, since each closed subset of $\mathbb{R}$ is determined by a countable rational basis for its open complement, and a countable sequence of such codes is again coded by a real. Under the Continuum hypothesis, enumerate all meagre $F_{\sigma }$ sets as $(M_{\alpha }{)}_{\alpha <{\omega }_1}$, allowing repetitions.

By transfinite recursion, chooseFor each $\alpha <{\omega }_1$, the excluded union is a meagre set: it is a countable union of meagre sets and singletons. The Baire category theorem ensures that it does not cover $\mathbb{R}$, so the choice is possible. The resulting $A$ is uncountable. If $M$ is meagre, choose $\gamma$ with $M\subseteq M_{\gamma }$. Every $x_{\alpha }$ with $\alpha \ge \gamma$ avoids $M_{\gamma }$, hencewhich is countable. Thus $A$ has precisely the Lusin set property. Enumerating covers rather than all meagre sets avoids an unjustified claim that there are only continuum many arbitrary meagre subsets.

The cardinal bound for an increasing chain of countable sets is $\boxed{|X|\le {\aleph }_1}$. If $X$ is countable, there is nothing to prove. Otherwise choose $Y\subseteq X$ with $|Y|={\aleph }_1$, and for every $y\in Y$ choose an index $i(y)$ such that $y\in X_{i(y)}$.

For any $i\in I$, its countable $X_i$ cannot contain all of $Y$, so choose $y\in Y\setminus X_i$. Since $I$ is a linear order, comparison of $i$ and $i(y)$ forces $i<i(y)$: the other direction would imply $y\in X_i$. Therefore $X_i\subseteq X_{i(y)}$. The selected family is cofinal among the original sets, andBy infinite cardinal arithmetic, this union of ${\aleph }_1$ countable sets has cardinality at most ${\aleph }_1$. No well-order or cofinality assumption on $I$ was used; a linear increasing union of countable sets has the same bound even for an arbitrary linear index order.

The continuum hypothesis holds. Let $A$ be the assumed Lusin set of cardinality $\mathfrak{c}=2^{{\aleph }_0}$. If $\mathfrak{c}>{\aleph }_1$, choose $B\subseteq A$ with $|B|={\aleph }_1$. The size assumption makes $B$ a meagre set, but then $A\cap B=B$ is uncountable, contradicting the Lusin set property. ThusOne can also use (d) directly: well-order $A$ in type $\mathfrak{c}$. Each proper initial segment has size below $\mathfrak{c}$, so is meagre and consequently countable, since it is a subset of the Lusin set. These initial segments form a linear increasing union of countable sets equal to $A$. The bound in (d) gives $\mathfrak{c}\le {\aleph }_1$.

Fix a regular uncountable cardinal number $\kappa$. A club set $C\subseteq \kappa$ is unbounded in $\kappa$ and closed under suprema below $\kappa$ of nonempty increasing sequences with no last term. Equivalently, every limit ordinal $\delta <\kappa$ with $\sup (C\cap \delta )=\delta$ belongs to $C$. The club filter isIt consists of all sets containing a club set, not only the club sets themselves.

An omega-measurable cardinal is a cardinal number carrying a nonprincipal ultrafilter closed under countable intersections, that is, a countably complete ultrafilter. Such a cardinal must be uncountable: on a countable set the intersection of the complements of all singletons would be empty.

A measurable cardinal is an uncountable cardinal number $\kappa$ carrying a nonprincipal $\kappa$-complete ultrafilter. A kappa-complete filter is closed under intersections of every family of size strictly below $\kappa$. Thus for an omega-measurable cardinal the completeness required is countable completeness, not merely closure under finitely many intersections.

A uniform example is $\boxed{A=\kappa \setminus \{\omega \}}$. It contains the final segment $[\omega +1,\kappa )$, a club set. Therefore $A$ belongs to the club filter. Every two club sets in a regular uncountable cardinal intersect: alternately choose larger points from each and take the countable supremum, which remains below $\kappa$ and lies in both by closure. Hence every member of the club filter is a stationary set, so $A$ is stationary.

However, every finite ordinal belongs to $A$, and their increasing supremum $\omega$ does not. Thus $A$ is not closed and is not a club set. This is a stationary nonclosed member of the club filter for every regular uncountable $\kappa$.

The club filter completeness argument works for every regular uncountable $\kappa$. Let $\mu <\kappa$ and let $C_i$ be a club set for each $i<\mu$. Their intersection is closed. To prove it unbounded, start above any prescribed $\beta <\kappa$ and choose an increasing sequence $({\delta }_n{)}_{n<\omega }$ so that ${\delta }_{n+1}$ lies above a chosen point of every $C_i$ greater than ${\delta }_n$. $\kappa$ is a regular cardinal, so the supremum of these $\mu$ choices remains below $\kappa$. The resulting countable supremum $\delta ={\sup }_n{\delta }_n$ likewise remains below $\kappa$. For each $i$, the chosen $C_i$ points are cofinal in $\delta$, so closure gives $\delta \in C_i$. Thus the intersection is a club set. Intersecting fewer than $\kappa$ members of the club filter still contains such an intersection of clubs. In particular, $\boxed{{\mathcal{D}}_{{\omega }_2}\text{ is }{\aleph }_2\text{-complete}}$.

It is not an ultrafilter. For a regular infinite $\theta <\kappa$, the setis stationary. Given a club set $C$, build in $C$ a strictly increasing continuous sequence of length $\theta$ and take its supremum $\delta <\kappa$. Closure gives $\delta \in C$, and the cofinality of an increasing ordinal supremum gives $\mathrm{cf}(\delta )=\theta$. This proves the stationarity of ordinals of prescribed cofinality.

At $\kappa ={\omega }_2$, the disjoint sets $S_{\omega }^{{\omega }_2}$ and $S_{{\omega }_1}^{{\omega }_2}$ are both stationary. A club contained in either $S_{\omega }^{{\omega }_2}$ or its complement would miss one of these stationary sets. Thus neither $S_{\omega }^{{\omega }_2}$ nor its complement belongs to ${\mathcal{D}}_{{\omega }_2}$, and

Weakly inaccessible cardinals are unbounded below the given cardinal. In fact, they form a stationary set there. Let $\kappa$ be the weakly Mahlo cardinal, so it is a regular uncountable limit cardinal and the set $R=\{\alpha <\kappa :\alpha =\mathrm{cf}(\alpha )\}$ is stationary.

The set $C$ of uncountable limit cardinals below $\kappa$ is a club set. For unboundedness, above any starting point choose a strictly increasing countable sequence of cardinal numbers below $\kappa$; its supremum remains below $\kappa$ by regularity and is an uncountable limit cardinal. For closure, a limit of such limit cardinals is again a limit cardinal.

Every $\alpha \in R\cap C$ is an uncountable regular cardinal and a limit cardinal, hence a weakly inaccessible cardinal. The intersection of a stationary set with a club set is stationary, because its intersection with any further club is nonempty. ThereforeThis proves the stronger form of the requested conclusion.

Take the least omega-measurable cardinal $\kappa$, with a countably complete ultrafilter $U$ on $\kappa$ that is nonprincipal. We prove that $U$ is $\kappa$-complete. This will give $\boxed{\kappa \text{ is measurable}}$.

Suppose $\mu <\kappa$, $A_i\in U$ for $i<\mu$, and ${\bigcap }_{i<\mu }{A}_i\notin U$. Then $B=\kappa \setminus {\bigcap }_{i<\mu }{A}_i$ belongs to $U$. For $x\in B$ let $g(x)$ be the least $i<\mu$ with $x\notin A_i$. The pushforward ultrafilteris a countably complete ultrafilter on $\mu$; this uses $B\in U$, so $U$ restricts to an ultrafilter on $B$. For each $i$, the fibre $g^{-1}[\{i\}]$ is contained in $\kappa \setminus A_i$, hence is not in $U$. Thus $W$ is nonprincipal. If $\mu$ is finite or countable, countable completeness already rules this out. Otherwise $|\mu |<\kappa$ would be an omega-measurable cardinal, contradicting minimality of $\kappa$.

Therefore every intersection of fewer than $\kappa$ members of $U$ lies in $U$. Since $\kappa$ is uncountable, it is a measurable cardinal. This proves that the least omega-measurable cardinal is measurable, even if the originally supplied cardinal carried only a countably complete ultrafilter.

The Von Neumann hierarchy and the constructible hierarchy are defined by transfinite recursion:The union clauses apply to nonzero limit ordinals. The definable power set $\mathrm{Def}(L_{\alpha })$ consists of subsets definable over $(L_{\alpha },\in )$ by a first-order formula with finitely many parameters from $L_{\alpha }$. Definability over that structure is essential; it is not unrestricted definability in the ambient universe. The full constructible universe is $L={\bigcup }_{\alpha \in \mathrm{Ord}}{L}_{\alpha }$.

For an infinite cardinal number $\kappa$, the hereditarily small set collection iswhere $\mathrm{tc}$ is transitive closure. Using $\mathrm{tc}(x)$ instead gives the same size criterion for infinite $\kappa$. This bounds the whole membership ancestry, not just $|x|$.

One can take $\boxed{\alpha =\omega +1}$. Finite induction gives $L_n=V_n$ for every finite $n$: every subset of a finite level is definable using its finitely many elements as parameters. Hence $L_{\omega }=V_{\omega }$, the set of hereditarily finite sets.

Now reason inside the constructible universe $L$, a model of ZFC. The set $L_{\omega }$ is countable there, and there are countably many first-order formulas and finite parameter tuples. Thus $L_{\omega +1}=\mathrm{Def}(L_{\omega })$ is countable in $L$. But the constructible power set ${\mathcal{P}}^L(\omega )$ is uncountable in $L$ by the Cantor theorem. ChooseThis $r$ is constructible and a subset of $\omega$, so $r\subseteq V_{\omega }$ and $r\in V_{\omega +1}$. ThereforeAll comparisons were made inside $L$; this avoids assuming that the constructible reals are uncountable in the ambient universe. The witness remains valid externally because the two structures use the same $r$, the same $V_{\omega }$, and the same constructible level. Thus constructible sets of low rank can appear at later stages.

Use the Lévy hierarchy, with bounded formulas in set theory as the quantifier-free level for this purpose. Let $\mathrm{Ord}(\kappa )$ be the bounded ordinal predicate, and let $\mathrm{Bij}(f,\alpha ,\kappa )$ be the bounded predicate saying that $f$ is a graph of a bijection from $\alpha$ onto $\kappa$. Then cardinalhood is expressed byAll quantifiers in $\mathrm{Ord}$ and $\mathrm{Bij}$ are bounded by the displayed sets and the graph. Ordered-pair membership, graph functionality, injectivity and surjectivity have bounded expansions using the usual set coding of ordered pairs. The only unbounded quantifier shown is universal over $f$. The formula is therefore ${\Pi }_1$ and in particular a Pi-one formula modulo ZF.

In ZF, a cardinal number is an ordinal not in bijection with any smaller ordinal, so this formula has exactly the required meaning. Thus cardinalhood is Pi-one definable; the argument does not use the axiom of choice.

Let $\mathrm{Map}(f,\delta ,\kappa )$ be the bounded graph predicate for a function $\delta \to \kappa$. For an infinite cardinal number, regularity is expressed byThe predicate $\mathrm{InfCard}$ is the cardinal predicate in (i), conjoined with the bounded condition that $\kappa$ is a nonzero limit ordinal. For an infinite cardinal number, this latter condition holds automatically; it excludes the finite cardinals. The expression $f(\xi )<\beta$ abbreviates bounded graph membership with the output lying in $\beta$.

Every quantifier after $\forall f$ is bounded, and the cardinal predicate is already ${\Pi }_1$. Their conjunction can be placed in universal form, so this is a Pi-one formula modulo ZF. Its mathematical content is that no function from a smaller ordinal is cofinal in $\kappa$. By the definition of cofinality, that is equivalent to $\mathrm{cf}(\kappa )=\kappa$. Hence regular cardinalhood is Pi-one definable. This uses the standard convention that a regular cardinal is infinite.

A witness to failure of the Axiom of power set is $\boxed{{\omega }_1\in H_{{\omega }_2}}$. Every subset $A\subseteq {\omega }_1$ belongs to $H_{{\omega }_2}$, since its transitive closure has cardinality at most ${\aleph }_1$. If some $p\in H_{{\omega }_2}$ were the internal power set of ${\omega }_1$, transitivity and bounded subsethood would force $p$ to contain every ambient subset of ${\omega }_1$, and only those subsets. Thus it would be the full $\mathcal{P}({\omega }_1)$.

By the Cantor theorem,Its transitive closure contains all its members, so $|\mathrm{tc}(\{p\})|\ge |p|\ge {\aleph }_2$. This contradicts $p\in H_{{\omega }_2}$. ThereforeThe power-set failure in hereditarily small sets needs no assumption about the Continuum hypothesis.

Let $T$ be ZFC together with the assertion of unboundedly many inaccessible cardinals. Given a model of $T$, if it has no inaccessible limit of inaccessibles, it already witnesses the required nonimplication. Otherwise let $\iota$ be its least inaccessible limit of inaccessibles and pass to its rank segment $V_{\iota }$.

Because $\iota$ is an inaccessible cardinal, $V_{\iota }$ satisfies ZFC. Its ordinals are precisely those below $\iota$, and below $\iota$ the inaccessible cardinals are unbounded by the choice of $\iota$. Inaccessibility of any smaller ordinal is absolute to this segment: it has the same subsets and functions on all smaller ordinals, as in strong-inaccessibility absoluteness from rank agreement. Thus $V_{\iota }$ satisfies the unbounded-inaccessibles assertion.

No smaller ordinal can be an inaccessible limit of inaccessibles, by minimality of $\iota$ and the same absoluteness. Also $\iota$ is not an ordinal of $V_{\iota }$ itself. ConsequentlyThis rank cutoff at the first inaccessible limit of inaccessibles proves $\mathrm{Con}(T)\Rightarrow \mathrm{Con}(T+\neg \mathrm{ILI})$. The soundness theorem for first-order logic then gives $T/⊢\mathrm{ILI}$ if $T$ is consistent. This is a model-theoretic relative-consistency argument, and does not infer external transitivity from mere consistency.

Use standard notation for forcing: $q\le p$ means that $q$ is stronger than $p$. A closed forcing order is $\lambda$-closed when every decreasing sequence $⟨p_{\xi }:\xi <\delta ⟩$ with $\delta <\lambda$ has a common lower bound:The chain condition for forcing isAn antichain in a forcing order consists of pairwise incompatible forcing conditions; incompatibility means having no common stronger extension. In particular, the ${\aleph }_1$-chain condition is the countable chain condition for forcing. These properties and their sequences or antichains are computed inside the ground model $M$.

False as printed: separativity alone does not guarantee a new generic filter. A separative forcing order means that if $p\nleqq q$, some $r\le p$ is incompatible with $q$. The one-point order $\mathbb{P}=\{p\}$ is separative vacuously. Its only generic filter is $\{p\}=\mathbb{P}$, and this belongs to $M$. Thus no $G\notin M$ exists in this example.

The natural missing hypothesis is that the nonempty order is an atomless forcing order. Under that hypothesis the intended argument works. Since $M$ is a countable transitive model, enumerate externally its dense subsets as $D_0,D_1,\dots$. Choose $p_{n+1}\le p_n$ with $p_{n+1}\in D_n$. Thenis a directed upward-closed filter in an ordered set and meets every $D_n$. This is the Rasiowa–Sikorski lemma, giving a generic filter over $M$.

If $G\in M$, the set difference $D=\mathbb{P}\setminus G$ would belong to $M$. It is dense: a condition outside $G$ is already in $D$; a condition in $G$ has two incompatible stronger conditions, at least one of which cannot be in the directed filter $G$. Genericity would require $G\cap D\ne \varnothing$, a contradiction. ThusThis uses generic filter for an atomless order is new. A forcing atom is precisely the obstruction illustrated by the counterexample; separativity does not remove it.

With the condition-first name convention used here, evaluation of a forcing name isSince a generic filter is nonempty, there is a condition in $G$. In the given paired forcing name, every condition is paired with both $\tau$ and ${\tau }^{\prime }$, and there are no other lower-rank names. ThereforeThis is the unordered pair of the two interpreted sets. If the interpretations coincide, it is a singleton. It is neither an ordered pair nor the set union of those interpretations. Reversing the two components in the name-pair convention changes only the syntax of the same recursive evaluation.

Closure preserves the ground-model cardinal $\lambda$. We use the stronger fact that closed forcing adds no short ordinal sequences. Fix an ordinal $\mu <\lambda$ in $M$ and a forcing name $\dot{h}$ that a condition $p$ forces to be an ordinal-valued function with domain $\mu$.

Inside $M$, recursively strengthen $p$ to decide each value $\dot{h}(\xi )$, for $\xi <\mu$. At a limit stage take a lower bound of the previously chosen descending sequence; its length is below $\lambda$, so $\lambda$-closure supplies one. After all $\mu$ values have been decided, closure supplies a common lower bound $q$. The decided values form a function $h\in M$ by the Axiom schema of replacement, andThe recursion is carried out within $M$ using the definable forcing theorem and a ground-model choice of deciding extensions, so every sequence to which closure is applied belongs to $M$. Repeating the construction below any stronger condition shows that such full-decision conditions are dense below a forcing condition $p$. The dense-below generic meeting lemma ensures that a generic filter containing $p$ meets them. Thus the interpreted function is in $M$.

If $\lambda$ ceased to be a cardinal number in $M[G]$, there would be a surjection $\mu \to \lambda$ for some ordinal $\mu <\lambda$. The forcing theorem gives a name and a condition in $G$ forcing this. The argument above makes its interpretation a ground-model function, but $M$ regards $\lambda$ as a cardinal number and admits no such surjection. ThereforeNo regularity of $\lambda$ is required; every recursion used has length strictly below $\lambda$. Finite cardinals are preserved automatically. This is cardinal preservation by closed forcing.

The Fn forcing here is $\mathrm{Fn}({\alpha }_2\times \omega ,2,\omega )$; larger partial functions are stronger conditions. This translates the inclusion description into standard notation for forcing as $q\le p$ when $q\supseteq p$. Compatible members of the generic filter agree wherever both are defined, sois a function. For every $(\xi ,n)\in {\alpha }_2\times \omega$, the conditions whose domains contain that coordinate form a dense subset of a forcing order. This requirement belongs to $M$, and the generic filter meets it. Consequently $F:{\alpha }_2\times \omega \to 2$ is total.

Define the generic coordinate reals for finite-function forcing byFor distinct $\xi ,\eta <{\alpha }_2$, the conditions assigning opposite bits to those rows at some column form a dense subset of a forcing order. Given any finite condition, choose a column absent from both rows and extend by assigning zero to one coordinate and one to the other. Genericity supplies such a condition in $G$, so $a_{\xi }\ne a_{\eta }$.

The function $\xi \mapsto a_{\xi }$ is formed inside $M[G]$ by separation and replacement from $F$. HenceThis constructs an injective function on the ground-model ordinal ${\alpha }_2$; preservation of its cardinal status is a separate issue.

Inside $M$, the finite-function order has the countable chain condition for finite-function forcing. In an uncountable family of binary conditions there are uncountably many distinct finite domains, since each domain supports only finitely many conditions. These domains would have an uncountable Delta-system subfamily by the Delta-system lemma. There are only finitely many assignments to the common finite root, so thin further to an uncountable family agreeing there. Any two such conditions have a union that is again a function and is a common stronger extension. Thus no uncountable antichain in a forcing order exists.

Here is the relevant preservation argument, rather than an appeal to the condition alone. Fix $p\in G$ forcing that a forcing name $\dot{h}$ is a function $\omega \to {\alpha }_1$. For each $n$, choose inside $M$ a maximal antichain in a forcing order below $p$ deciding $\dot{h}(n)$. By the countable chain condition for forcing, this antichain is countable in $M$, so the possible values form a countable $B_n\subseteq {\alpha }_1$. The union $B={\bigcup }_n{B}_n$ is countable in $M$, hence bounded below its regular ${\omega }_1={\alpha }_1$. The interpreted $h$ has range contained in $B$, by maximality and the dense-below generic meeting lemma. It therefore cannot be a surjection onto ${\alpha }_1$.

If ${\alpha }_1$ were not a cardinal number in $M[G]$, it would be equinumerous with a smaller ordinal, which was countable already in $M$; composing with that ground-model enumeration would give a surjection $\omega \to {\alpha }_1$. This is impossible. ThereforeThis is the possible-values lemma for chain-condition forcing specialized to ${\omega }_1$. Countability and regularity in this proof are computed inside $M$, not inferred from the external countability of $M$.