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

Let

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

$\\operatorname{cf}\\beta=\\gamma$ and $\\langle\\xi(\u):\u<\\gamma\\rangle$ 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 $\\gamma\eq\\beta$ . Then $\\gamma<\\beta$ by $\\operatorname{cf}\\delta\\leq\\delta$ .

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

单词	operatornamecfoperatornamecfalphaoperatornamecfalpha
释义	$\\operatorname{cf}(\\operatorname{cf}\\alpha)=\\operatorname{cf}\\alpha$ Let $\\operatorname{cf}\\alpha=\\beta$ and $\\langle\\alpha_{\\xi}:\\xi<\\beta\\rangle$ be cofinal in $\\alpha$ , and $\\operatorname{cf}\\beta=\\gamma$ and $\\langle\\xi(\u):\u<\\gamma\\rangle$ 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 $\\gamma\eq\\beta$ . Then $\\gamma<\\beta$ by $\\operatorname{cf}\\delta\\leq\\delta$ . Now, $\\langle\\alpha_{\\xi(\u)}:\u<\\gamma\\rangle$ is seen to be confinal in $\\alpha$ , which means that $\\operatorname{cf}\\alpha=\\gamma<\\beta$ , a contradiction. Therefore, $\\gamma=\\beta$ .
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cf⁡(cf⁡α)=cf⁡α

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