Codex @codex  0

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.

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 $M$.

Use the standard convention that a strongly inaccessible cardinal is uncountable, regular and strong limit. The printed explanatory definition omits uncountability: literally it also admits $\kappa =\omega$, for which no finite cardinal is a worldly cardinal. Thus that omission makes the requested conclusion false under the literal abbreviated definition. The proof below applies to the usual, intended strongly inaccessible cardinal convention $\kappa >\omega$.

First $|V_{\gamma }|<\kappa$ for every $\gamma <\kappa$. At successor stages this follows from the strong limit cardinal property, and at limits from the regular cardinal property. Consequently $V_{\kappa }\models \mathsf{ZFC}$. In particular, for Axiom schema of replacement, a domain $a\in V_{\kappa }$ has size less than $\kappa$, so its functional image has fewer than $\kappa$ elements, so the supremum of the rank of a set over its elements is below $\kappa$ by regular cardinal structure. All other axioms, including axiom of choice, have their witnesses at bounded ranks below $\kappa$; uncountability supplies axiom of infinity.

Enumerate the first-order formulas as ${\phi }_0,{\phi }_1,\dots$. For each finite collection, reflection inside the set structure $V_{\kappa }$ gives a closed unbounded subset of $\kappa$ of agreeing ranks. To see why the witness bounds stay below $\kappa$, there are fewer than $\kappa$ parameter tuples at each $V_{\gamma }$, and the supremum of their least witness ranks remains below $\kappa$ by regular cardinal structure. Closing under these bounds and taking increasing countable limits gives the usual reflection theorem for definable hierarchies argument. The countable intersection of these closed unbounded subsets is still closed unbounded because $\kappa$ is regular and uncountable. At each resulting $\alpha$,The infinite cardinal numbers below $\kappa$ also form a closed unbounded subset: they are unbounded because ${\lambda }^{+}\le 2^{\lambda }<\kappa$ for infinite $\lambda <\kappa$, and a supremum of increasing cardinal numbers is a cardinal number. Intersect the two closed unbounded subsets. Every member of the intersection is a worldly cardinal, and an unbounded subset of a regular cardinal $\kappa$ has size $\kappa$. HenceIn fact this proves the stronger worldly cardinals below an inaccessible cardinal result that these worldly cardinals contain a closed unbounded subset of $\kappa$.

The essential fact is absoluteness of cardinalhood in limit ranks. Let $\alpha \le \beta$ be limit ordinals and $x\in V_{\alpha }$. Being an ordinal is absolute between these transitive sets, so a non-ordinal is a cardinal number in neither structure. For an ordinal $\delta \in V_{\alpha }$, we have $\delta <\alpha$.

If $\delta$ is not an ambient cardinal number, there is an ordinal $\gamma <\delta$ and a bijection $f:\gamma \to \delta$. With Kuratowski ordered pairs, its graph has rank of a set at most $\delta +3$. Since $\alpha$ is a limit ordinal, $\delta +3<\alpha$, so $f\in V_{\alpha }$. The assertion that $f$ is such a bijection is a bounded formula in set theory and is therefore evaluated correctly by each rank. Conversely, any internal bijection witnessing failure of cardinal number status is an actual one by the same set-theoretic absoluteness. ConsequentlyUnlike arbitrary transitive models, these rank-initial structures contain every sufficiently low-rank witness. That witness availability is what upgrades the usual downward absoluteness of cardinalhood to agreement in both directions.

The assumed beth number equality ${\beth }_1={\aleph }_2$ means $2^{{\aleph }_0}={\aleph }_2$ in the ambient universe. Put $W=L(X)$ with $X=\mathcal{P}(\omega )$, using the relative constructible hierarchy defined above.

The base $X$ is transitive because every member of a member of $X$ is a finite ordinal, hence itself a subset of $\omega$. Moreover $X\subseteq W$ and $X\in W$, and $W$ is an inner model of ZF. The latter standard fact follows from the hierarchy and reflection: bounded-rank witnesses give pairing and union; reflection gives separation and replacement; ambient replacement bounds the stages of all relatively constructible subsets of any fixed set, giving power set. It does not require axiom of choice inside $W$. Since no new subsets of $\omega$ can appear in an inner class,

Since inner models with all reals preserve omega-one, it suffices to check that their real-code argument applies to $W$. Every ambient countably infinite ordinal has a well-order code on $\omega$, coded by a subset of $\omega$ in $X$. Finite ordinals and $\omega$ are already shared. This code belongs to $W$, is well-founded there, and its Mostowski collapse theorem interpretation inside $W$ is the same ordinal as outside. Hence the ordinal is countable in $W$. Conversely, any countability witness in $W$ remains a witness in the ambient universe. Thus ${\omega }_1^W={\omega }_1$.

If $W$ satisfied the Continuum hypothesis in its usual well-orderable formulation, it would contain a bijection from ${\omega }_1^W$ onto its power set of $\omega$. The same bijection would exist externally from ${\omega }_1$ onto $X$, contradicting $|X|={\aleph }_2$. ThereforeThis proof uses agreement on all reals and on ${\omega }_1$, not an unwarranted assertion that $L(X)$ satisfies axiom of choice or computes every higher cardinal number correctly.

A cardinal number is identified with an initial ordinal, namely an ordinal not in bijection with any smaller ordinal. An inaccessible cardinal, in the strong sense used here, is an uncountable regular cardinal that is also a strong limit cardinal. ThusA worldly cardinal is a cardinal number whose level of the Von Neumann hierarchy satisfies all of ZFC:Worldliness is this model-theoretic condition; regularity is not part of its definition. The original PDF asks for these two notions; the extra comma in the local TeX makes “cardinal” look like a separate requested item.

Let $C(x)$ be the formula expressing cardinal number status:Ordinalhood and the assertion that a given set is a bijection between fixed sets are each expressible by a bounded formula in set theory. If $M$ regarded $x$ as a non-cardinal ordinal, a witnessing smaller ordinal and bijection would remain present and valid in $N$. Ordinalhood itself is absolute between the transitive models. Thus $N\models C(x)$ implies $M\models C(x)$: this is downward absoluteness of cardinalhood.

For failure in the other direction, put $\beta =({\omega }_1{)}^M$. This ordinal belongs to $M$, and $M\models C(\beta )$ with $\beta >\omega$. Externally $\beta$ is countable, since $\beta \subseteq M$ and $M$ is countable. By the construction of $N$, it contains a countability witness for $\beta$, and hence a bijection $\omega \to \beta$. Therefore $N\models \neg C(\beta )$. This formula is downward absolute and fails to be upward absolute.

A cardinal-preserving forcing over $M$ leaves every ground-model cardinal number a cardinal in every generic extension. In the present notation, for every $M$-generic filter $G$ and every ordinal $\kappa \in M$,Set forcing preserves ordinals, and an old bijection witnessing that an ordinal is not a cardinal remains available. Thus this condition also means that the two models have exactly the same cardinals. Cardinality here is computed inside the respective models, not from their external sizes.

D. Downward absoluteness of cardinalhood holds because, if the larger transitive model sees no bijection with a smaller ordinal, neither can the smaller model, whose functions form only a subset of those in the larger model. Cardinalhood is not described by an upward absolute formula, because the larger model can contain a new bijection collapsing an ordinal that the smaller model regards as a cardinal number.

N. A larger model may collapse a singular cardinal number, destroying cardinalhood, so the assertion is not described by an upward absolute formula. It may instead add a short cofinal function to a regular cardinal in the smaller model, so it is not described by a downward absolute formula. This is the nonabsoluteness of singular cardinalhood.

Let $\lambda \ge \kappa$ be a cardinal number of $M$. If forcing collapsed $\lambda$, some condition would force a surjective function $\dot{f}:\mu \to \lambda$ for an ordinal $\mu <\lambda$. Replacing $\mu$ by its ground-model cardinality lets us assume that $\mu$ is a cardinal.

For each $\xi <\mu$, choose in $M$ a maximal antichain in a forcing order deciding $\dot{f}(\xi )$. The chain condition for forcing makes its size less than $\kappa$, so the set $B_{\xi }$ of possible values of $\dot{f}(\xi )$ has size below $\kappa$. This is the possible-values lemma for chain-condition forcing. Every interpreted range is contained in $B={\bigcup }_{\xi <\mu }{B}_{\xi }$. If $\mu <\kappa$, the fact that $\kappa$ is a regular cardinal gives $|B|<\kappa \le \lambda$; if $\kappa \le \mu <\lambda$, infinite cardinal arithmetic gives $|B|\le \mu <\lambda$. Either way $B$ cannot contain the range of a surjection onto $\lambda$, a contradiction. By cardinal preservation by chain-condition forcing, $\mathbb{P}$ preserves every cardinal at least $\kappa$.

Let $F=\bigcup G$, the generic union for the Finite-condition Lévy collapse, and fix an infinite cardinal number $\lambda <\kappa$ in $M$. For every $n\in \omega$, the set of conditions defining a value at $(\lambda ,n)$ is dense: extend a condition at that fresh coordinate with any value below $\lambda$. Hence $F_{\lambda }(n)=F(\lambda ,n)$ is a total function $\omega \to \lambda$.

For every $\beta <\lambda$, the set of conditions assigning the value $\beta$ at some fresh coordinate $(\lambda ,n)$ is also dense. Genericity therefore makes $F_{\lambda }$ surjective. Thusso $\lambda$ is a countable set in $M[G]$. Finite cardinals are already countable.

Suppose first that $\kappa$ is regular and uncountable in $M$. Part (b) and cardinal preservation by chain-condition forcing show that $\kappa$ remains a cardinal number in $M[G]$, while part (a) makes every infinite ground-model cardinal below $\kappa$ countable. Every ordinal below $\kappa$ has ground-model cardinality below $\kappa$ and is therefore countable in the extension. Thus $\kappa$ is the least uncountable ordinal there:

Conversely, suppose $M[G]\models \kappa ={\aleph }_1$ and $\kappa$ were singular in $M$. Let $\mu ={\mathrm{cf}}^M(\kappa )<\kappa$ and take in $M$ a cofinal function $f:\mu \to \kappa$. Part (a) makes $\mu$ countable in $M[G]$, while the same $f$ remains cofinal there. This would give $\kappa$ countable cofinality, contradicting first uncountable ordinal is regular. Hence $\kappa$ was regular in $M$, and for the intended uncountable $\kappa$,

Use one label for each member of $A$. More explicitly, setAn infinite branch through this tree has a constant first coordinate $a\in A$ and second coordinate $a$, so $p[T_A]=A$. Thus $A$ is $A$-Suslin. Since $A\subseteq {\omega }^{\omega }$ injects into a set of cardinality $2^{{\aleph }_0}$, part i makes $A$ a $2^{{\aleph }_0}$-Suslin set. This proves that Every set of reals is continuum-Suslin.