TITLE: Why cofinality is a cardinal.
QUESTION [7 upvotes]: 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?

REPLY [2 votes]: 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 $\gamma<\beta$.