I have seen two definitions for regular cardinals:
- A cardinal $\kappa>\omega$ is regular iff $\mathrm{cof}(\kappa)=\kappa$; that is, the cofinality of $\kappa$ is $\kappa$.
- For every map $f:\kappa\to\kappa$, there exists an ordinal $0<\alpha<\kappa$ that is closed under $f$; that is, $\forall \beta<\alpha(f(\beta)<\alpha)$.
I tried to prove that they are equivalent, but no luck. A cursory search on Google didn't turn up anything.
Yes, these are equivalent. The first property is the standard definition.
If $\operatorname{cf}(\kappa)=\kappa>\omega$, and $f\colon\kappa\to\kappa$, fix any $\alpha_0<\kappa$, and define by recursion, $\alpha_{n+1}=\sup\{f(\alpha)\mid\alpha\leq\alpha_n\}$. By regularity, $\alpha_{n+1}<\kappa$, and so $\alpha=\sup\alpha_n<\kappa$ as well.
In the other direction, if $\operatorname{cf}(\kappa)=\mu<\kappa$, consider a function $f\colon\mu\kappa$ which is cofinal and $f(0)=\mu$, and extend it by setting $f(\alpha)=0$ for all $\alpha\geq\mu$. It is easy to see that if $\alpha<\mu$, then $\alpha$ is not closed under $f$, since $f(0)>\alpha$; and on the other hand, if $\alpha\geq\mu$ then $\sup\{f(\beta)\mid\beta<\alpha\}=\kappa>\alpha$.