While working on a project in operator algebras with a [collaborator (and fellow MO user)](https://mathoverflow.net/users/10368/chris-heunen), we are able to successfully complete a transfinite induction assuming that the following has an affirmative answer.



> 
> **Question:** Let $\delta$ be a cardinal, considered as an [initial
>  ordinal](http://en.wikipedia.org/wiki/Ordinal_number#Initial_ordinal_of_a_cardinal), so that it is equal to the
>  set of all ordinals of cardinality
>  strictly less than $\delta$. Does
>  there exist a nondecreasing function
>  $\phi \colon \delta \to \delta$ such
>  that, for all ordinals $1 \leq \lambda < \delta$,
>  there exists $\gamma < \lambda$ such
>  that $\phi(\gamma) \geq \lambda$?
> 
> 
> 


If it happens to matter, we are only concerned with the case where $\delta$ is a limit cardinal. Any reasonably (transfinitely) constructive approach to writing down such $\phi$ seems to quickly run into issues of [ordinal notation](http://en.wikipedia.org/wiki/Ordinal_notation) that are beyond our expertise. If an abstract existence argument is available, we will certainly still be happy. On the other hand, we fear that this may somehow depend on large cardinal issues. 



Does this "jumping-ahead" ordinal function exist?
Not unless $\delta$ has countable cofinality (e.g., $\delta = \aleph\_\omega$). This will fail for $\delta = \aleph\_1$, for example. Let $\phi: \delta \to \delta$ be any increasing function and recursively define $\lambda\_0 = 0$ and $\lambda\_{n+1} = \phi(\lambda\_n)+1$. Since $\phi$ is increasing, the sequence $(\lambda\_n)$ is increasing, and since $\delta$ has uncountable cofinality we have $\lambda = \sup \lambda\_n < \delta$. However, for any $\gamma < \lambda$ we must have $\gamma < \lambda\_n$ for some $n$,
so that $\phi(\gamma) \leq \lambda\_{n+1} < \lambda$.


(If $\delta$ has countable cofinality it's easy. For instance, if $\delta = \aleph\_\omega$ then we define $\phi$ by letting $\phi(\lambda) = \aleph\_{n+1} + \lambda$ for $\aleph\_n \leq \lambda < \aleph\_{n+1}$.)


