Self-functions of the class of ordinals growing fast Let $\lambda$ be a regular ordinal and let $S:=\{\beta:\beta<\lambda\}$. I wonder if it is possible to find a (non-decreasing, but this should not be a problem) function
$$
f\colon S\to S
$$
such that, for any $\alpha\in S$, there exists $\beta\in S$ such that
$$
\beta<\alpha< f(\beta) (<\lambda).
$$

Sorry for not explaining the context. The fact is that I am attempting a construction involving (homotopy) limits of shape an ordinal, by transfinite induction. For the moment, I know how to proceed when an ordinal is a successor or when it is the sup of a chain (shorter than it is) of smaller ordinals. To handle also the regular case it would be enough to have a function like the above, but I do not know if such a thing can exist. 

There is no such function; it is a corollary of the Fodor's lemma. Assume that there is a such $f$, so for each $\xi<\lambda$ we can choose $\beta_\xi< \xi$ such that $\xi<f(\beta_\xi)$. The function $g(\xi) = \beta_\xi$ is regressive, and the set $\lambda$ is stationary.
Hence, there is a stationary subset $S\subseteq \lambda$ such that $g$ restricting to $S$ is a constant function. Note that any stationary subset is unbounded in $\lambda$ (that is, for each $\alpha<\lambda$ we can choose $\gamma\in S$ greater than $\alpha$.) However, $\xi < f(g(\xi))$ holds for all $\xi \in S$, and $f(g(\xi))$ is constant, a contradiction.

I should mention that your function, if it exists, is ill-defined when $\alpha=0$. However, my proof works even if we exclude some finitely many points from a domain of $f$ (and in fact the domain of $f$ shouldn't be a stationary subset of $\lambda$.)
