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

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	$\operatorname{cf}(\operatorname{cf}\alpha)=\operatorname{cf}\alpha$
Canonical name	operatornamecfoperatornamecfalphaoperatornamecfalpha
Date of creation	2013-03-22 18:11:22
Last modified on	2013-03-22 18:11:22
Owner	yesitis (13730)
Last modified by	yesitis (13730)
Numerical id	8
Author	yesitis (13730)
Entry type	Proof
Classification	msc 03E04

cf⁡(cf⁡α)=cf⁡α

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