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.

  • $\begingroup$ 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

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.


It seems that the answer is no. Barney showed (can be found in Foreman's chapter in the handbook, Chapter 3, the section regarding Uniformization Ideal) that it is consistent that $\mathrm{Unif}_\omega \subsetneq \mathrm{Unif}_2$, where $S\in \mathrm{Unif}_2$ if there exists a ladder system on $S$, such that any 2-coloring of the ladder can be uniformized. He showed these are normal countably complete ideals (so contains all non-stationary sets). Hence any $S \in \mathrm{Unif}_2 - \mathrm{Unif}_\omega$ will be stationary and a counter-example.


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