# A quick question about proof of Bukovský-Hechler

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 $\mu = \sum_{\alpha < \kappa} |\alpha|^\lambda \color{red}{\le} |\alpha_0|^\lambda \cdot \kappa$?

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? – Mario Carneiro Dec 17 '12 at 15:37
@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 $\kappa$. – Rudy the Reindeer Dec 17 '12 at 15:43
I think that $\sum_{x < \kappa} 1 = \kappa$ and hence if $|\alpha|^\lambda = |\alpha_0|^\lambda$ then $\displaystyle \sum_{\alpha < \kappa}|\alpha|^\lambda = \sum_{\alpha < \kappa}|\alpha_0|^\lambda =|\alpha_0|^\lambda \sum_{\alpha < \kappa}1 = |\alpha_0|^\lambda \cdot \kappa$. – Rudy the Reindeer Dec 17 '12 at 15:45
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.) – Mario Carneiro Dec 17 '12 at 16:22
@MarioCarneiro While you're right about successor ordinals, it isn't true that all limit ordinals have cofinality $\omega$. Try googling "regular cardinal". – Miha Habič Dec 17 '12 at 20:48