# Cofinality of a limit ordinal is equal to the cofinality of a given normal operation - Proof

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 $S$).

• What is meant by "normal operation"? What is the relationship between $\lambda$ and $\alpha$? – Jalex Stark May 8 '18 at 12:37
• Normal operation means the operation is both monotone and continuous and what do you mean the relationship between them? – AustereTiger May 8 '18 at 12:43
• That's it! Thank you. I meant to write $cf t_{\lambda}$. – AustereTiger May 8 '18 at 12:58
• Okay. Well as a starting hint, you need to show both that the left-hand-side is $\leq$ the right-hand side, and that the right-hand-side is $\leq$ the left-hand-side. Which one seems easier to start with? – Jalex Stark May 8 '18 at 12:59
• The left hand side, I suppose? – AustereTiger May 8 '18 at 15:26

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 $\mathrm{cf}(\lambda) \le \mathrm{cf}(t_\lambda)$.