Singular cardinal enumeration (${\sigma }_{\alpha }$)

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

There is an uncountable cardinal number $\kappa$ with ${\sigma }_{\kappa }=\kappa$. Iterate ${\kappa }_{n+1}={\sigma }_{{\kappa }_n}$ starting with ${\aleph }_0$. If there is no earlier fixed point, the increasing supremum $\kappa$ has countable cofinality, hence is a singular cardinal. The singular cardinal enumeration is cofinal in $\kappa$ below index $\kappa$, and continuity at this singular supremum gives equality.