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Put $\theta =\mathrm{cf}(\delta )$. There is a strictly increasing cofinal map $a:\theta \to \delta$. If $\mathrm{cf}(\theta )=\nu <\theta$, choose a cofinal map $b:\nu \to \theta$. Then $a\circ b$ is cofinal in $\delta$, contradicting the definition of $\theta$ as its least cofinal order type. Since always $\mathrm{cf}(\theta )\le \theta$, we obtainFor a nonzero limit ordinal this is an infinite regular cardinal. If the convention includes zero among limit ordinals, its cofinality is zero and the identity is immediate there.

For a nonzero limit ordinal $\delta$, a subset $C\subseteq \delta$ is a club set if it is unbounded in $\delta$ and contains every one of its limit points below $\delta$. A subset $S\subseteq \delta$ is stationary ifThis defines stationarity in a limit ordinal. The regular uncountable case is the usual stationary set setting; results such as Fodor lemma require that additional hypothesis, rather than an arbitrary limit ordinal.

For uncountable regular $\kappa$, let $\theta =\mathrm{otp}(C_{\zeta })$. Since $C_{\zeta }$ is a club set in $\zeta$, $\mathrm{cf}(\theta )=\mathrm{cf}(\zeta )=\kappa$, so $\theta \ge \kappa$.

If $\theta >\kappa$, its cofinality ensures $\theta >\kappa +\omega$: ordinals strictly between $\kappa$ and $\kappa +\omega$ are successors, and $\kappa +\omega$ has cofinality $\omega$. In the continuous increasing enumeration $c$ of $C_{\zeta }$, take $\gamma =c(\kappa +\omega )$. It is a limit point of $C_{\zeta }$ and has cofinality $\omega <\kappa$. Coherence gives $C_{\gamma }=C_{\zeta }\cap \gamma$, with order type $\kappa +\omega$ and size $\kappa$. This contradicts the size clause for ordinals of cofinality below $\kappa$. ThusThe printed hypotheses need this qualification. At $\kappa =\omega$, choose $C_{\zeta }=\zeta$ for every nonzero countable limit ordinal. These club sets are coherent, and the small-cofinality clause is vacuous. But at $\zeta ={\omega }^2$ their order type is ${\omega }^2$, not $\omega$. The standard square principle includes an order-type bound $\mathrm{otp}(C_{\zeta })\le \kappa$, which also repairs this countable case.

Fix $\delta <{\kappa }^{+}$ with $\mathrm{cf}(\delta )>\omega$. The limit points of $C_{\delta }$ form a club set in $\delta$. For each such point $\gamma$, coherence givesAs $\gamma$ increases through those limit points, these order types strictly increase. Hence the club set contains at most one member of a fixed fiber $E_{\xi }$. Removing an initial segment past that member leaves a club set disjoint from $E_{\xi }$. Therefore $E_{\xi }\cap \delta$ is not stationary. This proves each $E_{\xi }$ is non-reflecting, by the nonreflection of fixed order-type fibers.

For uncountable regular $\kappa$, part (a) gives type $\kappa$ at cofinality $\kappa$, and the size clause gives type less than $\kappa$ at smaller cofinalities. All nonzero limit ordinals below ${\kappa }^{+}$ fall into one of these cases. ThusThe index set $\kappa +1$ has cardinality $\kappa$, so this is a disjoint union of $\kappa$ non-reflecting subsets. Empty fibers can be omitted; if nonempty pieces are required, a fiber of size ${\kappa }^{+}$ can be split further, since nonreflection is inherited by subsets. Such a fiber exists because $\kappa$ sets of size at most $\kappa$ cannot cover ${\kappa }^{+}$ points.

At $\kappa =\omega$, the nonreflection definition is vacuous for every subset of ${\omega }_1$, since no smaller ordinal has uncountable cofinality. The partition conclusion is therefore still true by an arbitrary countable partition, although the particular fibers from the weak printed square assumptions need not cover all limit ordinals. Under the standard order-type-bounded square definition, the displayed fiber partition works in this case too.

The axiom of infinity excludes $V_{\omega }$ and all smaller stages, so the cardinal $\lambda$ is uncountable. Full semantics for second-order logic is decisive: every externally specified functional relation whose ordered pairs lie in $V_{\lambda }$ is an allowed class parameter, even if the relation itself is not an element of $V_{\lambda }$.

If $\mathrm{cf}(\lambda )=\mu <\lambda$, choose an external cofinal function $h:\mu \to \lambda$. Its graph is an allowed class parameter: each input, output and ordered pair has rank of a set below the infinite cardinal $\lambda$, which is a limit ordinal. The domain $\mu$ belongs to $V_{\lambda }$. The second-order form of Axiom schema of replacement would give its range as an element of $V_{\lambda }$. But a cofinal subset of $\lambda$ has rank of a set $\lambda$ and cannot belong to $V_{\lambda }$. Therefore $\mathrm{cf}(\lambda )=\lambda$.

If $2^{\theta }\ge \lambda$ for a cardinal $\theta <\lambda$, choice supplies an external surjection $q:\mathcal{P}(\theta )\to \lambda$. The domain belongs to $V_{\lambda }$ and the ordered pairs of its graph have ranks below $\lambda$. Applying Axiom schema of replacement to this external functional class would put its range, the ordinal $\lambda$, in $V_{\lambda }$, again impossible. Hence $2^{\theta }<\lambda$ for every $\theta <\lambda$.

ConsequentlyThese are exactly the defining conditions for a strongly inaccessible cardinal. This full second-order replacement rank obstruction would not follow from Henkin semantics, where the available class relations may exclude the external functions just used.

The stationary-indexed club principle ${♣}_S$ asserts a sequence $⟨A_{\delta }:\delta \in S\cap \mathrm{Lim}({\omega }_1)⟩$ with each $A_{\delta }\subset \delta$ cofinal of order type $\omega$, such that for every uncountable $X\subseteq {\omega }_1$ some $\delta \in S$ satisfies $A_{\delta }\subseteq X$. It predicts a cofinal ladder contained in the target, rather than predicting the entire initial segment of the target as diamond does. For a stationary $S$ with successor members, only its stationary limit-ordinal part is relevant; countable limit ordinals have cofinality $\omega$. This is the restricted version of the club principle.

A tree with unique limits has no distinct nodes at a limit level with the same history. Precisely, for a limit ordinal $\delta$ and $s,t\in T_{\delta }$,Here $s\upharpoonright \beta$ denotes the unique predecessor of $s$ of height $\beta$. This is uniqueness of a limit node when it exists, not a requirement that every cofinal chain in a partial order below a limit level acquire a limit node.

A kappa-filtration is an increasing continuous sequence with union $A$ and $|A_{\alpha }|<\kappa$ at every stage. Intersect $S$ with the club set of nonzero limit ordinals. For each remaining $\alpha$, continuity gives $A_{\alpha }={\bigcup }_{\beta <\alpha }{A}_{\beta }$, so choose $r(\alpha )<\alpha$ with $f(\alpha )\in A_{r(\alpha )}$. This is a regressive function. Fodor lemma gives a stationary subset $S_1$ and a fixed $\beta$ such that all these values lie in $A_{\beta }$.

Since $|A_{\beta }|<\kappa$, partition $S_1$ into fewer than $\kappa$ fibers of $f$. If every fiber were nonstationary, choose a club set avoiding each one. Their intersection is club set by regularity, contradicting stationarity of $S_1$. Thus one fiber is stationary, andThis proves the filtration form of Fodor lemma; continuity at limit stages, the small size of each stage, and regularity of $\kappa$ are all used.

The recursive definition of ordinal addition is $\alpha +0=\alpha$, $\alpha +(\beta +1)=(\alpha +\beta )+1$, and $\alpha +\lambda ={\sup }_{\beta <\lambda }(\alpha +\beta )$ for a nonzero limit ordinal $\lambda$. The synthetic definition is the order type of the ordered disjoint union of a copy of $\alpha$ followed by a copy of $\beta$.

For the synthetic operation, adding the empty order does nothing, adding a successor appends one final point, and concatenation with a limit order is the increasing union of the initial concatenations with all its proper initial segments. Their order types have supremum ${\sup }_{\beta <\lambda }(\alpha +\beta )$. Thus the synthetic operation satisfies the recursive equations. Transfinite induction gives uniqueness of the recursively defined operation, proving equivalence.

The recursive definitions of ordinal multiplication and ordinal exponentiation areThe last equation in each line is for nonzero limit $\lambda$. In particular the convention $0^0=1$ belongs to this recursive definition.

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.