# Preserve unbounded sets between different cofinality

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 $$k$$

• 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. Apr 28 '20 at 16:33
• @JohannesSchürz doesn't similar thing works for $\kappa$ weakly compact? Regardless on $λ$
– Holo
Apr 28 '20 at 17:23
• @HanulJeon but we are looking at the closure of the image, not just the image itself
– Holo
Apr 29 '20 at 10:03
• "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
– Holo
Apr 29 '20 at 11:21
• @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 $>\lambda$)
– 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.
• 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 $G_i$