An infinite cardinal $\kappa$ is regular if $\mathrm{cf}(\kappa) = \kappa$.
It is known that if $\kappa$ is regular, then for any family $(\kappa_i)_{i \in I}$ of cardinals $\kappa_i < \kappa$ with $|I| < \kappa$ $\kappa \neq \sum_{i \in I} \kappa_i$.
I want to prove the converse: let $\kappa$ be an infinite cardinal; if for any family $(\kappa_i)_{i\in I}$ of cardinals $\kappa_i < \kappa$ with $|I| < \kappa$ we have $\kappa \neq \sum_{i \in I} \kappa_i$, then $\kappa$ is regular.
One should possbily proceed by contradiction. Assume that $\mathrm{cf}(\kappa) < \kappa$. Then there is an ordinal $\alpha < \kappa$ for which there is a cofinal map $f\colon\alpha\to\kappa$, that is, a map $f\colon \alpha\to\kappa$ so that $(\forall \beta < \kappa)(\exists \gamma < \alpha)(\beta \leq f(\gamma))$. From this we should somehow derive a contradicting by showing that $\kappa$ is equal to $\sum_{i \in I} \kappa_i$ for some family $(\kappa_i)_{i \in I}$ of cardinals.
Best Answer
There really is only one option that sticks out: $$\sum_{\gamma\in\alpha}f(\gamma)$$