If $f\colon\kappa\rightarrow\kappa$, then the set of all $\alpha < \kappa$ such that $f(\xi) < \alpha$ for all $\xi < \alpha$ is closed and unbounded. If $f\colon\kappa\rightarrow \kappa$, then the set of all $\alpha < \kappa$ such that $f(\xi) < \alpha$ for all $\xi < \alpha$ is closed and unbounded.
This is from Jech's book (page 103) so I guess $\kappa$ is an ordinal or a cardinal. However, it seems to me that if $\kappa$ was an ordinal then the statement would not be true. For example, let $f(0)=f(1)=0$ and for each $n > 1$ let $f(n) = n-1$. Then for each $1 < n < \omega$ we have $f(n)<n$ but $f(\omega) = \omega$. Am I right?
I have found a proof in the internet, but it uses stationary sets and Fodor's Lemma. However, It does seems to me that if I could figure out weather $\kappa$ denotes an ordinal or a cardinal, then I will be able to find a proof for this, which uses more basic definitions, like the definitions of transitive sets an ordinals.
What do you think?
Thanks

Closed and unbounded sets are uninteresting in countable cofinality, so one has to at least assume that the cofinality of $\kappa$ is uncountable.
It's also not true otherwise, since if $\kappa=\delta+\omega$, then the function $f(\alpha)=\begin{cases}\alpha & \alpha<\delta \\ \alpha+1 & \delta\leq\alpha\end{cases}$ is a function from $\kappa$ to itself which doesn't satisfy this property.
So you should assume, at least that the cofinality of $\kappa$ is uncountable; but you don't have to assume that this is a cardinal. I'll give you a hint:


*

*Show that if for all $i\in I$, $\alpha_i$ satisfies this property, then $\alpha=\sup\alpha_i$ also satisfies this property, so this is a closed set.

*Show that given an ordinal $\xi$, we can find $\alpha>\xi$ with this property (use the fact that the cofinality is uncountable!) and therefore the set of such $\alpha$'s is unbounded.
