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The following is an exercise in Just/Weese (page 179),

Question 1: can you tell me if I got it right? Thank you!

Question 2: Shouldn't it be equality rather than less equals in $\mu = \sum_{\alpha < \kappa} |\alpha|^\lambda \color{red}{\le} |\alpha_0|^\lambda \cdot \kappa$?

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Assume $|\alpha_0|^\lambda < \kappa$. Then $|\alpha|^\lambda < \kappa$ for all $\alpha < \kappa$. Then $\displaystyle \sum_{\alpha < \kappa}|\alpha|^\lambda = |\alpha_0|^\lambda \cdot \kappa = \kappa$.

We know that if $\kappa$ is an infinite cardinal then $\kappa$ is singular if and only if there exists an $\alpha < \kappa$ and a set of cardinals $\{ \kappa_\xi : \xi < \alpha \}$ such that $\kappa_\xi < \kappa$ for all $\xi < \alpha$ and $\kappa = \sum_{\xi < \alpha} \kappa_\xi$.

Hence if $|\alpha_0|^\lambda < \kappa$, then $\kappa$ must be regular since there is no $\gamma < \kappa$ with $\sum_{\alpha < \gamma} |\alpha|^\lambda = |\alpha_0|^\lambda \cdot \gamma = \kappa$. But by assumption, $\kappa$ is singular.

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    It seems to me that the less-equal deduction comes from $\alpha\leq\alpha_0\Rightarrow|\alpha|^\lambda\leq|\alpha_0|^\lambda$ so that if $\alpha<\kappa$ implies $|\alpha|^\lambda=|\alpha_0|^\lambda$, then $|\alpha|^\lambda\leq|\alpha_0|^\lambda$ for all $\alpha<\kappa$, so $\sum_{\alpha<\kappa}|\alpha|^\lambda\leq\sum_{\alpha<\kappa}|\alpha_0|^\lambda=|\alpha_0|^\lambda\cdot\kappa$. Do $c$ and $f(\kappa)$ have definitions?2012-12-17
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    @MarioCarneiro $\mathrm{cf}(\kappa)$ is used to denote the cofinality of $\kappa$ and is defined to be the smallest ordinal $\delta$ such that there exists a function $f: \delta \to \kappa$ such that the range of $f$ is confinal in $\kappa$.2012-12-17
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    I think that $\sum_{x < \kappa} 1 = \kappa$ and hence if $|\alpha|^\lambda = |\alpha_0|^\lambda$ then $\displaystyle \sum_{\alpha < \kappa}|\alpha|^\lambda = \sum_{\alpha < \kappa}|\alpha_0|^\lambda =|\alpha_0|^\lambda \sum_{\alpha < \kappa}1 = |\alpha_0|^\lambda \cdot \kappa$.2012-12-17
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    But you only know $\alpha\geq\alpha_0\Rightarrow |\alpha|^\lambda=|\alpha_0|^\lambda$. For $\alpha<\alpha_0$, you can only conclude $|\alpha|^\lambda\leq|\alpha_0|^\lambda$. (viz. $\operatorname{cf}(\kappa)$: I hate it when people italicize multi-letter function names.)2012-12-17
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    @MarioCarneiro You are right! Thank you.2012-12-17
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    Correct me if I'm wrong, but cofinality seems trivial, i.e. $\alpha$ is cofinal with $\beta$ iff both are successor ordinals, unless one is 0, so $\operatorname{cf}(\kappa)=\omega$ if $\kappa$ is a limit ordinal and $\operatorname{cf}(\kappa)=1$ otherwise, although I think this last case is impossible for infinite cardinals.2012-12-17
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    @Mario: That last comment makes no sense. You should search for "cofinality" on this website search engine, there are some explanations how it works, but it is definitely not the way you wrote it.2012-12-17
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    @Asaf What I mean is that if $\alpha=\beta+1$ for some $\beta$, then the function $f:1\to\alpha$ given by $f(0)=\beta$ is a monotone ordinal function, so $1$ is cofinal with $\alpha$. Infinite cardinals are always limit ordinals, though, as $|\kappa|=|\kappa+1|$ if $\kappa\geq\omega$, by the bijection $f:\kappa+1\to\kappa$ given by $f(\kappa)=0$, $f(n)=n+1$, and $f(\alpha)=\alpha$ if $\omega\leq\alpha<\kappa$.2012-12-17
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    @MarioCarneiro While you're right about successor ordinals, it isn't true that all limit ordinals have cofinality $\omega$. Try googling "regular cardinal".2012-12-17

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