Working in ZFC, let $\kappa,λ$ be cardinals with $\kappa>λ$, and assume that $\kappa$ is regular.

We say that a function $F:\kappa^n→λ$, for some finite $n$, is preserving unbound, if for all $a⊆\kappa$ unbounded, we have that the closure of $F[a^n]$ under $F$ is unbounded in $λ$.

*Is there always preserving unbound function?*

For $\mbox{cof}(λ)=ω$ this is easy: take $n=1$ and map every element from $\kappa\setminusλ$ into some some element of $λ$, and map an element from $λ$ to somewhere greater using some fixed cofinal sequence of $λ$, i.e. fix some cofinal $(λ_i\mid i\in\omega)$, and for $k$ be the minimal $k$ such that $x∈λ_k$, map $x$ to some element of $\lambda_{k+1}\setminusλ_k$.

The problem arise when $\mbox{cof}(λ)>ω$. In this case we cannot "climb" a cofinal.

In this case for $n=1$ there are clearly no preserving unbound function. Indeed if $F:\kappa→λ$ be any function, then there exists some $x∈λ$ such that $F^{-1}(x)$ is unbounded in $\kappa$.

A special case is that $λ$ is regular, in this case preserving unbound becomes:

A function $F:\kappa^n→λ$, for some finite $n$, such that if for all $a⊆\kappa$ unbounded, we have $F[a^n]$ is unbounded in $λ$.

Indeed if $F[a^n]$ is bound, then $|F[a^n]|<λ$, then $|\bigcup\{F[a^n],F[F[a^n]],...\}|<λ$, so the closure is also bounded.

The I am most interested in the special cases where $λ$ is indeed regular and that $\kappa=λ^+$, in particular $\kappa=ω_{k+1},λ=ω_k$ for finite $k$

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