In the proof of the lemma

For every limit ordinal $\alpha$, $\textrm{cf } \alpha$ is a regular cardinal

Jech says that

It is easy to see that if $\alpha$ is not a cardinal, then using a mapping of $|\alpha|$ onto $\alpha$ one can construct a cofinal sequence in $\alpha$ of length $\leq |\alpha|$, and therefore $\textrm{cf } \alpha < \alpha$.

I've spent a while trying to construct this cofinal sequence and was wondering if I was on the right path:

From a bijection $f$ between $\kappa = |\alpha|$ and $\alpha$ take $S = \{\beta \in \kappa : \ f(\gamma) < f(\beta)$ for all $\gamma < \beta\}$ then if $\xi$ is the order type of $S$ and $g$ is the isomorphism from $\xi$ to $S$ the function $f\circ g$ is a cofinal sequence in $\alpha$ and $\xi < \alpha$.

Is this on the right track, or is there an easier way of doing it?


1 Answer 1


Yes, this idea does work. In Jech's setting, you should check that $f\circ g$ is strictly increasing (which holds by construction of $S$ and definition of $g$), and that it has a cofinal range.

For this last item, it is enough to show that the range of $f$ is cofinal in $\alpha$, that is, for every $\delta\in\alpha$ there exists some $\beta_0\in S$ such that $f(\beta_0)\geq \delta$. And for this,

take $\beta_0 :=\min \{\beta \in \kappa : \delta \leq f(\beta)\}$.

Then by construction $f(\gamma) < f(\beta)$ for all $\gamma<\beta$.

  • 1
    $\begingroup$ Awesome thanks, do you happen to know a simpler way of doing this or is this the normal way of doing it. $\endgroup$
    – Kai Rüsch
    Feb 10, 2017 at 15:31
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    $\begingroup$ You're welcome. I think something more standard would be to do some recursive definition of the cofinal sequence, out of elements of $\kappa$. That would be like computing $g$ and proving that the composition is cofinal at the same time. Another treatment of this subject is in Kunen's The Foundations of Mathematics, Theorem I.13.11. $\endgroup$ Feb 10, 2017 at 15:33

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