I've been going through Hrbacek and Jech's Introduction to Set theory, 3rd edition, on my own. There's an exercise which states:

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

I think I proved the opposite, $\kappa=\aleph_\omega$ is a counter example, since all the cardinals below it are successor cardinals, besides $\aleph_0$, and successor cardinals are regular.

Is my counterexample correct?

In detail: Consider $\kappa=\aleph_\omega$, a limit cardinal. It has cofinality $\omega$. Consider an increasing sequence of cardinals, $\langle\alpha_n \;|\; n<\omega\rangle$ with limit $\kappa$. $\aleph_0$ is regular, and every cardinal between $\aleph_0$ and $\aleph_\omega$ is a successor cardinal, and therefore regular. Hence for all $n$, $cf(\alpha_n)=\alpha_n$, which means there cannot be a single $\lambda$ equal to the cofinality of every element, the sequence is increasing. $\lambda=\aleph_1<\kappa$ is a concrete example of a regular infinite cardinal for which this statement fails.


  • $\begingroup$ You get a vote for finding a logical error in a textbook, even if it may only be a typo. $\endgroup$ Aug 22, 2017 at 18:14

1 Answer 1


You are correct. The statement should assume that $\lambda<cf(\kappa)$, instead of that $\lambda<\kappa$. This fixes your counterexample, since for $\kappa=\aleph_\omega$ there is no infinite regular cardinal $\lambda<cf(\kappa)$.

  • $\begingroup$ This might be a typo in the book...............+1 $\endgroup$ Aug 22, 2017 at 18:13
  • $\begingroup$ I thought it might be a typo, but I wasn't sure what the typo would be. Thanks. $\endgroup$ Aug 22, 2017 at 18:50

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