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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 $p$.

We construct the two transitive models together, taking care that the later collapse really retains the earlier model. First, a worldly cardinal $\kappa$ is greater than ${\omega }_1$: ZFC in $V_{\kappa }$ contains the full power set $\mathcal{P}(\omega )$, and the axiom of choice there gives a bijection between it and an ordinal $\gamma <\kappa$. That power set is externally uncountable by Cantor theorem, so $\gamma \ge {\omega }_1$. Infinity rules out the finite cardinals and $\omega$ before this argument is applied.

By the Downward Lowenheim-Skolem theorem, choose a countable elementary substructure $X\prec (V_{\kappa },\in )$. Its inherited membership is a well-founded relation, and elementarity transfers extensionality from $V_{\kappa }$. Apply the Mostowski collapse theorem to obtain a countable transitive model $M$ of ZFC. Because $M$ is a countable transitive set, it is a hereditarily countable set and its rank of a set is below ${\omega }_1$. In particular, $M\in V_{\kappa }$.

Now take the Skolem hull in $V_{\kappa }$ of the seedCall it $H$. The seed has external cardinality ${\aleph }_1$ and lies inside $V_{\kappa }$. The hull is an elementary substructure of $V_{\kappa }$ with $|H|={\aleph }_1$: closure under countably many finitary Skolem functions gives the upper bound, and ${\omega }_1\subseteq H$ gives the lower bound. Let $\pi :H\to N$ be its Mostowski collapse theorem map. Then $N$ is a transitive model of ZFC.

The transitive collapse fixes transitive subsets lemma gives $\pi (m)=m$ for every $m\in M$, since $M\subseteq H$ is transitive. Since also $M\in H$, the collapse sends that element to $M$ itself. Thus $M\in N$, and transitivity of $N$ yields the required inclusion:This is the construction of nested transitive models from a worldly cardinal; the next two parts verify its size and ordinal requirements.

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.

U+D. By absoluteness of infinitude between transitive models, two transitive models of ZFC agree on the natural numbers and therefore on whether a shared set is a finite set or an infinite set.

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.

U+D. The assertion that a given relation is a partial order quantifies only over its underlying set and checks that it is a reflexive relation, an antisymmetric relation and a transitive relation. It is therefore a bounded formula in set theory and is absolute between transitive models.

U+D. The natural numbers are absolute between transitive models of ZFC, and the statement $x\subseteq \mathbb{N}$ is the bounded assertion $\forall y\in x(y\in \mathbb{N})$. Hence the property of being a subset of $\mathbb{N}$ is absolute.

Let $M$ be the given transitive model and suppose that its ordinal height of a model of set theory $\mathrm{Ord}\cap M$ were a countable set. For every $x\in M$, the internal Axiom of choice gives a bijection from $x$ to an ordinal of $M$; transitivity makes this an actual bijection, and the ordinal is externally countable. Thus every element of $M$ is externally countable.

For each $\alpha \in \mathrm{Ord}\cap M$, the internal rank $V_{\alpha }^M$ belongs to $M$ and is therefore countable. Every $x\in M$ lies in one of these ranks, so $M$ is a countable union of countable sets and is itself countable, contrary to the hypothesis. By uncountable transitive set model has uncountable ordinal height, $M$ contains uncountably many ordinals.

Fix $a\in L$. Inside the ambient transitive model $M$, the Axiom of power set makes the collection of constructible subsets of $a$ a set. For each such subset $x$, choose the least stage of the constructible hierarchy at which $x$ appears. The Axiom schema of replacement and the supremum of a set of ordinals give an ordinal $\beta$ bounding all these stages; enlarge $\beta$ so that $a\in L_{\beta }$.

Nowis definable over $L_{\beta }$ with parameter $a$. It therefore belongs to the definable power set $\mathcal{D}(L_{\beta })=L_{\beta +1}\subseteq L$. This set contains exactly the subsets of $a$ that belong to the constructible universe, so it witnesses the Axiom of power set in $L$. Therefore $L\models$ Power Set.

Take a transitive model $M\models \mathsf{ZFC}+2^{{\aleph }_0}={\aleph }_2$. In $M$, choose a bijection $e:{\omega }_2\to \mathcal{P}(\omega )$ and encode its graph by a set $A\subseteq {\omega }_2$, using a fixed bijection between ${\omega }_2\times \omega$ and ${\omega }_2$. The relative constructible universe $L(A)$ can decode $e$, and therefore contains every real number of $M$; being an inner model of $M$, it has no additional reals.

Models with the same reals have the same first uncountable ordinal, because their reals code exactly the same countable well-orders. If $L(A)$ satisfied the Continuum hypothesis, its bijection between ${\omega }_1$ and the reals would also belong to $M$, contradicting $M\models 2^{{\aleph }_0}={\aleph }_2$. This is the construction in relative constructible universe can violate the continuum hypothesis, and it gives

A syntactically ${\Pi }_1$ first-order formula has an unbounded universal quantifier block followed by a bounded formula in set theory. Universal quantification makes it downward absolute between transitive models. A Pi-one formula modulo ZF may have another syntax but has such an equivalent provable in ZF.

A first-order formula is ${\Pi }_1^{\mathrm{ZF}}$ if ZF proves it equivalent, with the same free variables, to a universal unbounded quantifier block followed by a bounded formula in set theory. Bounded quantifiers do not increase its level in the Lévy hierarchy. Such properties are downward absolute between appropriate transitive models of ZF.

Assume a transitive model $M$ satisfies $2^{{\aleph }_0}={\aleph }_2$. In $M$, encode a bijection $e:{\omega }_2\to \mathcal{P}(\omega )$ by one set $A\subseteq {\omega }_2$, using a fixed pairing of ${\omega }_2\times \omega$ with ${\omega }_2$. Then $L(A)$ decodes $e$ and contains every real of $M$. The two models consequently have the same ${\omega }_1$, and any bijection between ${\omega }_1$ and the real numbers in $L(A)$ would also be one in $M$. Thus $L(A)\models \neg \mathsf{CH}$.

A first-order formula is syntactically ${\Sigma }_1$ when it is an unbounded existential quantifier block followed by a bounded formula in set theory. Equivalence provable in a specified theory gives the corresponding modulo-theory notion. Existential witnesses show upward absoluteness between transitive models containing the parameters.

A model whose relation symbol for membership is interpreted by actual membership restricted to its domain. If the domain is also transitive, it is a transitive model. Some conventions include this transitivity requirement in the word standard.

A standard model of set theory interprets membership by the ambient membership relation restricted to its domain. A transitive model additionally has a transitive set or class as its domain. Standardness alone does not impose transitivity.

A first-order formula is ZF-equivalent bounded if ZF proves it equivalent, with the same free variables, to a syntactically bounded formula in set theory. It need not itself have only bounded quantifiers. Between transitive models of ZF, use the bounded equivalent and the axioms in each model to prove set-theoretic absoluteness. The syntactic bounded-formula result requires only transitivity; the provable-equivalence extension additionally requires the theory used for the equivalence.