# Cofinal sequence with uniform cofinality - Hrbacek and Jech, Introduction to Set Theory, Chapter 9, exercise 2.6

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).

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 $$|S|=cf(\kappa).$$

• Well, arguably there are no regular cardinals below $\omega$. At least if you understand the concept of regularity as something that involved infinitude... Aug 7 '21 at 0:55

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.)

• 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.
– jjs
Aug 7 '21 at 15:47
• (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 $\aleph_\alpha>\aleph_{\omega_1}$. Aug 7 '21 at 17:31