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Let $S \subseteq \omega_1$ be a stationary set of limit ordinals and let $L = \langle L_\alpha \;|\; \alpha\in S\rangle$ be a ladder system. We say that $L$ has $\kappa$-uniformization if for every sequence of functions $f_\alpha : L_\alpha\to \kappa$ ($\alpha\in S$), there is a function $F : \omega_1\to\kappa$ such that for every $\alpha\in S$, $F\upharpoonright L_\alpha =^* f_\alpha$, i.e. $F$ and $f_\alpha$ agree on a cofinite subset of $L_\alpha$.

It is known (see Eklof-Mekler-Shelah, Theorem 15) that, if $S$ is stationary and every ladder system on $S$ has $2$-uniformization, then in fact every ladder system on $S$ has $\omega$-uniformization.

Question: Suppose $S$ is stationary and $L$ is a ladder system on $S$ which has $2$-uniformization. Does it follow that $L$ has $\omega$-uniformization?


Edit: Since it seems the answer to the above question might be a little far off at the moment, let me ask a weaker question that I'm also interested in:

Question: Suppose $S$ is stationary and $L$ is a ladder system on $S$ which has $2$-uniformization. Let $f_\alpha : L_\alpha\to \omega$ be the collapse map. Does there exist an $F : \omega_1\to \omega$ such that for all $\alpha\in S$, $F\upharpoonright L_\alpha =^* f_\alpha$?

Note that, given $\diamondsuit(S)$, there exists a ladder system on $S$ such that the above "canonical" $\omega$-coloring (i.e. the coloring given by the collapse maps) has no uniformization; hence if the answer is yes, then it must come somehow from the $2$-uniformization property of the ladder system.

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  • $\begingroup$ Just to be clear, what exactly does a ladder system mean? $\endgroup$ – Asaf Karagila Feb 9 '15 at 23:44
  • $\begingroup$ @AsafKaragila: A ladder $L_\alpha$ in $\alpha$ is an increasing $\omega$-sequence of ordinals which is cofinal in $\alpha$. A ladder system on $S$ is just a sequence of ladders $L_\alpha$, one for each $\alpha\in S$. $\endgroup$ – Paul McKenney Feb 10 '15 at 0:37
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For partial progress, let me argue at least that $2$-uniformization implies $n$-uniformization for any finite $n$.

The idea will be to guess the binary digits of the desired function separately, and then put these values together.

Suppose we are given a system of functions $f_\alpha:L_\alpha\to n$ on the fixed ladder system $L_\alpha$ for $\alpha\in S$, where $S$ is stationary. Choose $k$ large enough so that $n\leq 2^k$, and for each $i<k$ let $f_\alpha^i(x)$ be the $i^{th}$ binary digit of $f_\alpha(x)$. Since $f_\alpha^i:L_\alpha\to 2$, we get by $2$-uniformization a function $F_i:\omega_1\to 2$ that almost threads the functions $f_\alpha^i$. Thus, $F_i(\beta)$ correctly provides the $i^{th}$ binary digit of $f_\alpha(\beta)$ for almost all $\beta\in L_\alpha$. Combining these individual digit functions into one function $F:\omega_1\to 2^k$, let $F(\beta)$ be the number whose $i^{th}$ binary digit is $F_i(\beta)$, for $i<k$. Since the functions $F_i$ almost-thread the functions $f_\alpha^i$, it follows that $F$ eventually has all the right binary digits as $f_\alpha\upharpoonright L_\alpha$, and so it almost threads the original functions $f_\alpha$.

I'm not sure if this idea can be used to establish $\omega$-uniformization, since it isn't sufficiently clear to me whether we can ensure that the digits stabilize quickly enough.

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