Let

$\operatorname{cf} \alpha=\beta$ and $\langle\alpha_\xi:\xi<\beta\rangle$ be cofinal in $\alpha$ , and

$\operatorname{cf} \beta=\gamma$ and $\langle\xi(\nu):\nu<\gamma\rangle$ be cofinal in $\beta$ .

The claim of the theorem $\operatorname{cf}(\operatorname{cf} \alpha) = \operatorname{cf} \alpha$ means that $\gamma=\beta$ ; we prove this fact.

Suppose $\gamma\neq\beta$ . Then $\gamma<\beta$ by $\operatorname{cf}\delta\leq\delta$ .

Now, $\langle\alpha_{\xi(\nu)}:\nu<\gamma\rangle$ is seen to be confinal in $\alpha$ , which means that $\operatorname{cf}\alpha=\gamma<\beta$ , a contradiction. Therefore, $\gamma=\beta$ .