# Possible typo in proof of Bukovský-Hechler

In the proof of the following theorem:

Theorem 29 (Bukovský-Hechler): Let $\kappa, \lambda$ be infinite cardinals such that $\mathrm{cf}(\kappa) \le \lambda$ and $\mathrm{cf}(\kappa) < \kappa$. Denote $\displaystyle \sum_{\alpha < \kappa} |\alpha|^\lambda = \mu$.

(a) If there exists $\alpha_0 < \kappa$ such that $|\alpha|^\lambda = |\alpha_0|^\lambda$ for all $\alpha$ satisfying $\alpha_0 \le \alpha < \kappa$, then $\kappa^\lambda = \mu$.

(b) If for each $\alpha < \kappa$ there exists $\beta$ such that $\alpha < \beta < \kappa$ and $\alpha^\lambda < \beta^\lambda$, then $\kappa^\lambda = \mu^{\mathrm{cf}(\mu)}$.

there appears to be a typo in the proof of part (b). Can you confirm this? The proof is the following:

I don't quite believe that $H$ maps into ${^{\mathrm{cf}(\kappa)}}F$. Assume $\gamma_\xi$ is somewhere between $\alpha < \gamma_\xi < \kappa$. Then $f(\beta)$ could also be somewhere above $\alpha$ so that $g \notin F$. I am not sure what am missing but I think the definition of $H$ should depend on $\alpha$. The fix I propose is the following:

$$g(\beta) = \begin{cases} f(\beta) & \gamma_\xi < \alpha \\ 0 & \text{otherwise}\end{cases}$$

-

I think you're misunderstanding the definition of $F$ (which is, admittedly, quite confusingly written, as is the whole proof). The $\alpha$ in the definition isn't free; $F$ is supposed to be the set of bounded functions from $\lambda$ to $\kappa$. It is clear that $H(f)(\xi)$ is bounded by $\gamma_\xi$, so the map $H$ is well defined, at least.

It is much better to imagine $H$ a bit more concretely. In particular, uncurrying the definition, we basically have a map $H\colon{\,}^\lambda\kappa\times\mathrm{cf}(\kappa)\to F$, which takes a function $f\in {}^\lambda\kappa$ and a point $\xi$, and produces a map $H(f,\xi)$ which is just $f$, cut off at the appropriate point $\gamma_\xi$ in the cofinal sequence.

-

$\newcommand{\cf}{\operatorname{cf}}$When you assume that $\gamma_\xi$ is between $\alpha$ and $\kappa$, you’re presupposing some fixed $\alpha$, but there isn’t a fixed $\alpha$ anywhere in this argument. The $\alpha$ in the definition of $F$ is just an index variable:

$$F=\bigcup_{\alpha<\kappa}{^\lambda\alpha}\;,$$

the set of all bounded functions from $\lambda$ into $\kappa$. Equivalently,

$$F=\{f\in{^\lambda\kappa}:\sup\operatorname{ran}f<\kappa\}\;.$$

It may make things a little clearer if I try to give a more intuitive description of $H$.

An element of $\,{^{\cf\kappa}F}$ is a function that assigns to each $\xi<\cf\kappa$ an $f_\xi\in F$, i.e., a bounded function from $\lambda$ into $\kappa$. In other words, it’s a $(\cf\kappa)$-sequence of bounded functions from $\lambda$ into $\kappa$. The function $H$ then assigns to each $f:\lambda\to\kappa$ a $(\cf\kappa)$-sequence of bounded functions from $\lambda$ into $\kappa$ in a very natural way.

We have the increasing cofinal sequence $\langle\gamma_\xi:\xi<\cf\kappa\rangle$ in $\kappa$, so every bounded function into $\kappa$ must be bounded by some $\gamma_\xi$. We set $H(f)=\langle f_\xi:\xi<\cf\kappa\rangle$, where $f_\xi$ is the result of zeroing out each value of $f$ that is greater than or equal to $\gamma_\xi$. That is, we set $f_\xi(\eta)=f(\eta)$ if $f(\eta)<\gamma_\xi$, but if $f(\eta)$ exceeds that bound, we set $f_\xi(\eta)=0$. As $\xi\to\cf\kappa$, $f_\xi$ agrees with $f$ on more and more values of the argument. In fact, for any $\eta<\lambda$ there is a $\xi_0<\cf\kappa$ such that $f(\eta)<\gamma_{\xi_0}$, and we have $f_\xi(\eta)=f(\eta)$ for $\xi_0\le\xi<\cf\kappa$, so that $$\lim_{\xi\to\cf\kappa}f_\xi(\eta)=f(\eta)$$ for all $\eta<\lambda$.

-