Consider a normal operation assigning $t_\alpha$ to each ordinal number $\alpha$, and assume that $\lambda$ is a limit ordinal.
Show that cf $t_\lambda$ $=$ cf $\lambda$.
I want to show that the smallest cardinal $\kappa$ such that $\lambda$ is the supremum of $\kappa$ smaller cardinals is equal to the cofinality of the operation. Help would be appreciated here. I know the cofinality of any ordinal $\alpha$ is the least cardinal number $\kappa$ such that there exists a subset $S$ of the ordinal $\alpha$ having cardinality $\kappa$ and that $\alpha$ is the least ordinal $\textit{strictly}$ greater than every member of $S$ (the strict supremum of $S$).
Sketch:
Pick a cofinal subset $S \subseteq \lambda$. Then $$ T = \{ t_s \mid s \in S \} $$ is cofinal in $t_\lambda$. This shows that $\mathrm{cf}(t_\lambda) \le \mathrm{cf}(\lambda)$.
Conversely, let $T \subseteq t_\lambda$ be cofinal. Then $$ S = \{ \alpha \mid t_\alpha \in T \} $$ is cofinal in $\lambda$. Hence $\mathrm{cf}(\lambda) \le \mathrm{cf}(t_\lambda)$.