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The exercise is as follows:

Let $\kappa$ be a limit cardinal, and let $\lambda < cf (\kappa)$ be a regular infinite cardinal. Show that there is an increasing sequence $\langle \alpha_{\nu} \mid \nu < cf (\kappa)\rangle$ of cardinals such that $\lim_{\nu \to cf(\kappa)} \alpha_{\nu} = \kappa$ and $cf(\alpha_{\nu}) = \lambda$ for all $\nu$.

Hrbacek and Jech in fact write "$\lambda < \kappa$" instead of "$\lambda < cf (\kappa)$", but there is a counterexample for this (as shown in Hrbacek and Jech, sequence with uniform cofinality, ch 9 exercise 2.6), so I take it to be a typo.

I can show that there is an increasing sequence of infinite cardinals cofinal in $\kappa$ and with the length of the sequence being $cf(\kappa)$. A sequence of cardinals $\langle \lambda, \aleph_{\lambda}, \aleph_{\aleph_{\lambda}}, \dots \rangle$ of cofinality $\lambda$ can also be obtained from the facts that $cf(\lambda) = \lambda$ and $cf(\aleph_{\lambda}) = cf(\lambda)$, since $\lambda$ is an infinite cardinal and hence a limit ordinal. Can these be reasonably combined to get the desired result? If not, I have no idea how to proceed and would appreciate a hint (not a complete solution, please).

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This is wrong if $cf(\kappa)=\omega,$ as the example $\kappa=\omega_{\omega}$ shows.

If $cf(\kappa)>\omega:$ For any ordinal $x$ let $f_{\lambda}(x)$ be the least cardinal $y>x$ such that the set of cardinals between $x$ and $y$ has cardinality $\lambda.$ Show $x<\kappa\implies f_{\lambda}(x)<\kappa.$ Recursively let $G(x+1)=f_{\lambda}(G(x))$ and if $x=\cup x$ then let $G(x)=f_{\lambda}(\cup_{u\in x}G(u)).$ Now take $S\subset \{G(x):x\in \kappa\}$ with $\cup S=\kappa$ and $|S|=cf(\kappa).$

A simpler approach is to show that $T=\{y<\kappa: |y|=y\land cf(y)=\lambda\}$ is unbounded in $\kappa,$ so let $S\subset T$ with $\cup S=\kappa$ and $|S|=cf(\kappa).$

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    $\begingroup$ Well, arguably there are no regular cardinals below $\omega$. At least if you understand the concept of regularity as something that involved infinitude... $\endgroup$
    – Asaf Karagila
    Aug 7 at 0:55
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Note that if $\mu<\kappa$, then $\mu^{+\lambda}<\kappa$, since any cardinal between $\mu$ and $\mu^{+\lambda}$ is either a successor or has cofinality of at most $\lambda$.

Pick any cofinal sequence, and apply the above observation.

(You were vaguely on the right path, but $\lambda\mapsto\aleph_\lambda$ is a function that jumps over way too many cardinals, way too fast. So it's not going to work in all cases.)

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  • $\begingroup$ I see that the ordinal exponentiation can be used to define normal ordinal operations, which preserve cofinality. If I could show what you state above, I could then use the operations to transform a cardinal sequence cofinal in $\kappa$ to an ordinal sequence of the same length, still cofinal in $\kappa$, with $cf(\alpha_{\nu}) = \lambda$ for all $\nu$ in the domain of the sequence. What I do not see are (1) how to show that limit cardinals between $\mu$ and ${\mu^{+}}^{\lambda}$ have cofinality at most $\lambda$, and (2) how to obtain a cardinal sequence from the ordinal sequence. $\endgroup$
    – jjs
    Aug 7 at 15:47
  • $\begingroup$ (1) It's an interval of cardinals, so any cardinal in that gap is of the form $\mu^{+\alpha}$ for some $\alpha\leq\lambda$. But $\lambda$ is regular, so for limit ordinals, which correspond to the limit cardinals, the cofinality of $\alpha$ is smaller, and therefore the cofinality of $\mu^{+\alpha}$ is smaller. (2) Why should you be able to obtain such a sequence? Consider $\aleph_{\omega_1}$, take a cofinal sequence of ordinals of countable cofinality, almost all of these $\alpha$s satisfy $\aleph_\alpha>\aleph_{\omega_1}$. $\endgroup$
    – Asaf Karagila
    Aug 7 at 17:31

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