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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    $\begingroup$ If $\kappa$ is measurable and $\lambda$ regular such a function cannot exist. Assume the contrary. W.l.o.g. let $n=2$. Then we can define the function $G(\{\alpha,\beta\}):=\max(F((\alpha,\beta)), F((\beta,\alpha)))$. But there exists a measure 1 set $a$ such that $G\restriction [a]^2$ is constant. Therefore, $F$ cannot be preserving unbounded. $\endgroup$ Apr 28 '20 at 16:33
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    $\begingroup$ @JohannesSchürz doesn't similar thing works for $\kappa$ weakly compact? Regardless on $λ$ $\endgroup$
    – Holo
    Apr 28 '20 at 17:23
  • $\begingroup$ @HanulJeon but we are looking at the closure of the image, not just the image itself $\endgroup$
    – Holo
    Apr 29 '20 at 10:03
  • $\begingroup$ "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 $\lambda_{k+1}\setminusλ_k$" If $\kappa=ω_{ω+1}$ and $λ=ω_ω$, send all elements in $\kappa\setminusλ$ to some $x$. If $x\inω$, then $F(x)∈ω_1$, if $x∈ω_1$ then $F(x)∈ω_2$ etc. Then the closure of every subset of $λ$ under $F$ is cofinal $\endgroup$
    – Holo
    Apr 29 '20 at 11:21
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    $\begingroup$ @HanulJeon $F(\xi)$ is a fixed value, but the closure of $F[\kappa\setminus\lambda]$ under $F$ will be $\bigcup\{F(\xi),F(F(\xi)),...\}$, which will be unbounded(where $\xi$ is some arbitrary value $>\lambda$) $\endgroup$
    – Holo
    Apr 29 '20 at 11:41

In the case that $\kappa=\lambda^+$, I claim that $n=2$ suffices to produce such a function, as follows. Fix, for each ordinal $\beta<\kappa$ a one-to-one function $f_\beta:\beta\to\lambda$. Then define $F:\kappa^2\to\lambda$ by setting $F(\alpha,\beta)=f_\beta(\alpha)$ if $\alpha<\beta$. (It doesn't matter how you define $F(\alpha,\beta)$ for $\alpha\geq\beta$.) Now consider any unbounded $A\subseteq\kappa$; I want to show that $F(A^2)$ is unbounded in $\lambda$, for which it suffices to show that $F(A^2)$ has cardinality $\lambda$. Since $A$ is unbounded in $\kappa$, it has a $\lambda$-th element $\beta$. As $\alpha$ ranges over the $\lambda$ members of $A$ that are $<\beta$, $F(\alpha,\beta)=f_\beta(\alpha)$ takes $\lambda$ distinct values, because $f_\beta$ is one-to-one. And all of these $\lambda$ values are in $F(A^2)$ because $\beta$ and all the $\alpha$'s under consideration are in $A$.

I'm reasonably sure a similar argument (iterating this idea) will show that $n=q+1$ suffices when $\kappa$ is the $q$-fold successor of $\lambda$. Unfortunately, I don't have time right now to write down the argument.

  • $\begingroup$ Great! As for $\kappa=\lambda^{+q}$, $n=2$ is enough, if $F_i$ is such map for $(\lambda^{+i+1})^2\to\lambda^{+i}$, define $G_i$ to be $G_i(x,y)=(F_i(x,y),F_i(x,y))$, the. Compose all of $G_i$ $\endgroup$
    – Holo
    Apr 28 '20 at 16:34

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