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.

Question

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

Answer 1

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:

1. 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.

2. 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.