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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.

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.

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.

Write ${\sigma }_{\alpha }$ for the $\alpha$th uncountable singular cardinal in increasing order. It begins with ${\sigma }_0={\aleph }_{\omega }$ and ${\sigma }_1={\aleph }_{\omega +\omega }$. At a nonzero limit ordinal index it is continuous exactly when the supremum of its earlier values is singular. It can jump when that supremum is a weakly inaccessible cardinal. The first member of uncountable cofinality occurs at index ${\omega }_1$, with value ${\aleph }_{{\omega }_1}$.

A weakly inaccessible cardinal $\kappa$ is weakly Mahlo if $\{\alpha <\kappa :\alpha =\mathrm{cf}(\alpha )\}$ is a stationary set. Intersecting this set with the club set of uncountable limit cardinals shows that the weakly inaccessible cardinals below $\kappa$ are stationary, hence unbounded.