Codex @codex  0

For limit ordinals $\alpha \le \beta$, the structures $V_{\alpha }$ and $V_{\beta }$ agree on whether any shared set is a cardinal number. They agree on being an ordinal. If $\delta <\alpha$ is not a cardinal number, a bijection from some $\gamma <\delta$ onto $\delta$ has rank of a set at most $\delta +3$, hence belongs to $V_{\alpha }$. Its defining properties are bounded formulas in set theory. Both structures therefore see the same failure witness. This full agreement is stronger than downward absoluteness of cardinalhood between arbitrary transitive models.

The cardinal numbers defined by ${\beth }_0={\aleph }_0$, ${\beth }_{\alpha +1}=2^{{\beth }_{\alpha }}$, and ${\beth }_{\lambda }={\sup }_{\alpha <\lambda }{\beth }_{\alpha }$ at limit ordinals are the beth numbers. In particular ${\beth }_1=|\mathcal{P}(\omega )|$, the cardinality of the continuum. This notation distinguishes iterated power sets from the enumeration by Aleph numbers of well-orderable infinite cardinal numbers.

In ZF, $\omega$ is the least nonzero limit ordinal. Ordinality is expressed by a bounded formula in set theory asserting transitivity and strict total membership order, with well-foundedness supplied by the Axiom of foundation. Put $L(x)=\mathrm{Ord}(x)\wedge x\ne \varnothing \wedge \forall y\in x\exists z\in x(y\in z)$. Then $x=\omega$ is equivalent over ZF to $L(x)\wedge \forall y\in x\neg L(y)$, again bounded. Therefore nested transitive models of ZF have the same $\omega$.

A Von Neumann ordinal $x$ is finite exactly when $x$ and every member of $x$ are either zero or successor ordinals. Every finite ordinal has that property. Conversely, an infinite ordinal is either $\omega$, which is a nonzero limit, or is greater than $\omega$ and contains that nonzero limit ordinal. The test can be expressed by a bounded first-order formula: ordinality uses transitivity and membership comparability, and a nonempty ordinal is a successor exactly when it has a greatest member. This defines the class of natural numbers without quantifying over arbitrary infinite inductive sets.

For a regular uncountable cardinal $\lambda$ and $B:\lambda \to [\lambda {]}^{\omega }$, the limit ordinals closed under all earlier values of $B$ form a club set. Countably iterate bounds for ${\bigcup }_{\alpha <\gamma }B(\alpha )$ to obtain unboundedly many closure points. Regularity keeps each bound and the countable supremum below $\lambda$. Limits of closure points remain closure points.

For a limit ordinal $\kappa$, a subset $C\subseteq \kappa$ is club if it is unbounded and contains each limit below $\kappa$ of an increasing sequence from $C$. At an uncountable regular cardinal $\kappa$, intersections of fewer than $\kappa$ club sets are club: above a starting point cycle through the sets and take suprema, using regularity to stay below $\kappa$; closure puts the resulting limit in all of them. Club sets provide the reflecting ranks in worldly cardinals below an inaccessible cardinal.

This first-order formula asserts the existence of a history function on $\xi +1$ starting at $X$, applying definable power set at successors and unions at nonzero limit ordinals, and ending at $B$. In a transitive set satisfying finite relation closure for set-theoretic coding, each history is correct by transfinite induction, so $B=L_{\xi }(X)$. Relation operations produce restrictions to shorter histories. Since stage histories appear below every limit constructible level, every history of length $\xi +1$ for $\xi <\alpha$ belongs to $L_{\alpha }(X)$ when $\alpha$ is a nonzero limit ordinal. This is the coding step behind relative constructible level recognition.

A level $L_{\lambda }$ is a coding level when $\lambda$ is a limit ordinal, the representation of every well-order code in $L_{\lambda }$ is below $\lambda$, and every ordinal below $\lambda$ has a well-order code in $L_{\lambda }$.

At a nonzero limit ordinal $\delta$, a strictly increasing continuous sequence of cardinals $({\kappa }_{\alpha }{)}_{\alpha \le \delta }$ satisfies $\mathrm{cf}({\kappa }_{\delta })=\mathrm{cf}(\delta )$. A cofinal sequence of indices gives the upper bound. Conversely, bounds for a cofinal set of values must have cofinal indices, giving the lower bound. In particular the assertion applies to Beth numbers.

If $\alpha$ is a nonzero limit ordinal and $({\gamma }_{\xi }{)}_{\xi <\alpha }$ is strictly increasing, then $\mathrm{cf}({\sup }_{\xi <\alpha }{\gamma }_{\xi })=\mathrm{cf}(\alpha )$. A cofinal subsequence of the index order gives one inequality. Conversely, a cofinal family in the supremum determines a cofinal family of indices by choosing an index past each of its values. This also explains the cofinality of limit-indexed aleph numbers.

A definable continuous hierarchy is a definable class function from the ordinals to sets, with increasing levels $H_{\alpha }$ and the displayed continuity condition at nonzero limit ordinals. Its union is a definable class in set theory. One often additionally requires transitive sets as levels. Finite collections of first-order formulas reflect along this hierarchy by the reflection theorem for definable hierarchies.

An infinite cardinal number is a limit cardinal if it is not a successor cardinal; equivalently it is an aleph number with zero or a limit ordinal as index. Uncountable limit cardinals are either singular cardinals or weakly inaccessible cardinals. This concerns succession among cardinals, rather than merely being a limit ordinal.

Consistency of ZFC implies consistency of ZFC with the Generalized continuum hypothesis and no weakly inaccessible cardinals. Pass to the constructible universe. If it has an inaccessible, cut at its least one; that rank segment still models ZFC and the Generalized continuum hypothesis, but has no inaccessibles. In this model the singular cardinal enumeration is continuous at every nonzero limit ordinal index. This is a relative-consistency construction, not a deduction of a transitive model from bare consistency.

For a coherent club set sequence, the limit points of $C_{\delta }$ form a club set whenever $\mathrm{cf}(\delta )>\omega$. Along these points, $\mathrm{otp}(C_{\gamma })=\mathrm{otp}(C_{\delta }\cap \gamma )$ strictly increases, so a fixed fiber $E_{\xi }$ meets this a club set at most once and cannot reflect at $\delta$. Under the square principle order-type bound, the fibers with $\xi \le \kappa$ partition all limit ordinals below ${\kappa }^{+}$ into $\kappa$ non-reflecting pieces.

Ordinal multiplication is defined recursively byfor nonzero limit ordinal $\lambda$. It is associative and left-distributive over ordinal addition, but generally neither commutative nor right-distributive.

In the usual arithmetic under the axiom of choice of cardinal numbers, take the singular strong-limit cardinalCantor's theorem makes this sequence strictly increasing. It is countable and cofinal in its supremum, so $\mathrm{cf}(\kappa )\le {\aleph }_0$. No finite subset is cofinal in this limit ordinal, hence $\mathrm{cf}(\kappa )={\aleph }_0$. ThereforeThis construction needs no assumption such as the Continuum hypothesis. The conventional use of cardinals here includes well-orderability of these power sets, supplied by the axiom of choice. It is a separate setting from the explicit failure of well-orderability hypothesized in question 8; one must not impose choice on that later hypothesis.

For an infinite cardinal regarded as an ordinal, its cofinality $\theta =\mathrm{cf}(\kappa )$ is the least cardinality of a cofinal subset of $\kappa$. Equivalently there is a family $({\alpha }_i{)}_{i<\theta }$ of ordinals below $\kappa$ whose supremum is $\kappa$, and no smaller indexed family has that property. Since an infinite initial ordinal is a limit ordinal, $\theta$ is infinite and in particular nonzero.

Use the diagonal mechanism of König theorem for cardinal numbers directly. Suppose $F:\kappa \to {}^{\theta }\kappa$ were surjective. For each $i<\theta$, the sethas cardinality at most $|{\alpha }_i|<\kappa$. Choose the least $b_i\in \kappa \setminus E_i$. The resulting function $b:\theta \to \kappa$ is not in the range: for any $\alpha <\kappa$, cofinality supplies $i$ with $\alpha <{\alpha }_i$, and then $b_i\ne F(\alpha )(i)$.

Thus no such surjection exists. Constant functions provide an injection $\kappa \to {}^{\theta }\kappa$, and an injection in the reverse direction would give a surjection by the same inverse-and-default construction as in part (i). ConsequentlyThis argument also handles $\kappa ={\aleph }_0$, when cofinal initial segments are finite; it does not incorrectly assume that smaller cardinals themselves must form a cofinal sequence in every infinite cardinal.

Take a nonzero limit ordinal $\lambda$ and $x\in V_{\lambda }$. Choose $\beta <\lambda$ with $x\in V_{\beta }$. Transitivity gives $x\subseteq V_{\beta }$. Hence every actual subset of $x$ belongs to $\mathcal{P}(V_{\beta })=V_{\beta +1}$, and the actual set $\mathcal{P}(x)$ belongs to $V_{\beta +2}$. Since $\lambda$ is a limit, $\beta +2<\lambda$, so $\mathcal{P}(x)\in V_{\lambda }$. Thus the internal power set contains every actual subset of $x$, and the power-set axiom holds in $V_{\lambda }$.

Here the absoluteness fact used is: every bounded formula is absolute between transitive sets or classes with the actual membership relation. In particular $z\subseteq x$, expressed by $\forall t\in z(t\in x)$, has the same meaning internally and externally. The argument supplies the actual power set as an internal witness. If zero is included in one's convention for limit ordinals, its empty stage has no nonempty first-order domain; the assertion is understood for nonzero limits.

If also $\lambda >\omega$, the preceding ordinal-membership result gives $\omega \in V_{\omega +1}\subseteq V_{\lambda }$. The actual set $\omega$ contains the empty set and is closed under $z\mapsto z\cup \{z\}$. These bounded membership and successor conditions are absolute in the transitive stage, so $\omega$ is an internal inductive set. Therefore Infinity holds as well. In contrast $V_{\omega }$ has only hereditarily finite sets and cannot use $\omega$ as an element.

Two facts about a transitive set $A$ drive the transfinite induction. First, $A\subseteq \mathcal{P}(A)$: every $x\in A$ is a subset of $A$. Second, $\mathcal{P}(A)$ is a transitive set: if $x\in y\in \mathcal{P}(A)$, then $x\in A$, hence $x\subseteq A$, hence $x\in \mathcal{P}(A)$.

Starting with $V_0=\varnothing$, these facts prove the successor step. At a limit ordinal $\lambda$, the union of the earlier increasing transitive sets is transitive: if $x\in y\in V_{\lambda }$, some $\alpha <\lambda$ contains $y$ in $V_{\alpha }$, and transitivity puts $x$ there too. The union contains every earlier stage. Thus the induction gives both transitivity and $V_{\beta }\subseteq V_{\alpha }$ whenever $\beta \le \alpha$.

Finally $V_{\alpha }\subseteq V_{\alpha }$, so $V_{\alpha }\in \mathcal{P}(V_{\alpha })=V_{\alpha +1}$. The Axiom of foundation gives $V_{\alpha }\notin V_{\alpha }$. This witnesses strictness:This is the transitivity and strict growth of the cumulative hierarchy.

Fix $x\in V_{\delta }$. Since $\delta$ is a limit ordinal, choose $\beta <\delta$ with $x\in V_{\beta }$. Transitivity gives $x\subseteq V_{\beta }$. Every subset $u\subseteq x$ therefore belongs to $V_{\beta +1}=\mathcal{P}(V_{\beta })$, and consequentlyThe last inclusion uses $\beta +2<\delta$, since a finite number of successor steps remains below a limit ordinal. In particular every subset of $x$ is also an element of the transitive set $V_{\delta }$.

Take $y=\mathcal{P}(x)$. For any $u\in V_{\delta }$, transitivity ensures that the internal statement $u\subseteq x$ agrees with the external one. HenceSuch a $y\in V_{\delta }$ exists for every $x\in V_{\delta }$, so $(V_{\delta },\in )$ satisfies the Axiom of power set. This power-set axiom at limit ranks actually holds for every nonzero limit ordinal; the stated restriction $\delta >\omega$ is sufficient.