 Solution

The Beth numbers form a strictly increasing continuous sequence: strictness comes from Cantor theorem, and at a limit $\delta$ one has ${\beth }_{\delta }={\sup }_{\xi <\delta }{\beth }_{\xi }$. A cofinal sequence in $\delta$ of length $\mathrm{cf}(\delta )$ therefore induces a cofinal sequence in ${\beth }_{\delta }$, giving $\mathrm{cf}({\beth }_{\delta })\le \mathrm{cf}(\delta )$.

Conversely, let $\{a_i:i<\mathrm{cf}({\beth }_{\delta })\}$ be cofinal in ${\beth }_{\delta }$. For each $i$ choose ${\xi }_i<\delta$ with $a_i<{\beth }_{{\xi }_i}$. The indices must be cofinal in $\delta$; otherwise all $a_i$ would be bounded by a single smaller Beth number. Thus $\mathrm{cf}(\delta )\le \mathrm{cf}({\beth }_{\delta })$, andThis is the cofinality of a continuous cardinal hierarchy at a nonzero limit index; the limit hypothesis is essential to the supremum argument.