# $\operatorname{cf}(\operatorname{cf}\alpha)=\operatorname{cf}\alpha$

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, $\gamma=\beta$.

Title $\operatorname{cf}(\operatorname{cf}\alpha)=\operatorname{cf}\alpha$ operatornamecfoperatornamecfalphaoperatornamecfalpha 2013-03-22 18:11:22 2013-03-22 18:11:22 yesitis (13730) yesitis (13730) 8 yesitis (13730) Proof msc 03E04