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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 ${\omega }_1<{\aleph }_{{\omega }_1}$.

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.

The club filter completeness argument works for every regular uncountable $\kappa$. Let $\mu <\kappa$ and let $C_i$ be a club set for each $i<\mu$. Their intersection is closed. To prove it unbounded, start above any prescribed $\beta <\kappa$ and choose an increasing sequence $({\delta }_n{)}_{n<\omega }$ so that ${\delta }_{n+1}$ lies above a chosen point of every $C_i$ greater than ${\delta }_n$. $\kappa$ is a regular cardinal, so the supremum of these $\mu$ choices remains below $\kappa$. The resulting countable supremum $\delta ={\sup }_n{\delta }_n$ likewise remains below $\kappa$. For each $i$, the chosen $C_i$ points are cofinal in $\delta$, so closure gives $\delta \in C_i$. Thus the intersection is a club set. Intersecting fewer than $\kappa$ members of the club filter still contains such an intersection of clubs. In particular, $\boxed{{\mathcal{D}}_{{\omega }_2}\text{ is }{\aleph }_2\text{-complete}}$.

It is not an ultrafilter. For a regular infinite $\theta <\kappa$, the setis stationary. Given a club set $C$, build in $C$ a strictly increasing continuous sequence of length $\theta$ and take its supremum $\delta <\kappa$. Closure gives $\delta \in C$, and the cofinality of an increasing ordinal supremum gives $\mathrm{cf}(\delta )=\theta$. This proves the stationarity of ordinals of prescribed cofinality.

At $\kappa ={\omega }_2$, the disjoint sets $S_{\omega }^{{\omega }_2}$ and $S_{{\omega }_1}^{{\omega }_2}$ are both stationary. A club contained in either $S_{\omega }^{{\omega }_2}$ or its complement would miss one of these stationary sets. Thus neither $S_{\omega }^{{\omega }_2}$ nor its complement belongs to ${\mathcal{D}}_{{\omega }_2}$, and

For a regular infinite $\theta <\kappa$, with $\kappa$ a regular uncountable cardinal number, $S_{\theta }^{\kappa }=\{\delta <\kappa :\mathrm{cf}(\delta )=\theta \}$ is stationary. Build a strictly increasing continuous $\theta$-sequence in any club set, and take its supremum. Closure places it in that club and the cofinality of an increasing ordinal supremum gives cofinality $\theta$. In particular the two disjoint sets $S_{\omega }^{{\omega }_2}$ and $S_{{\omega }_1}^{{\omega }_2}$ show that the club filter on ${\omega }_2$ is not an ultrafilter.