*This is a bit of an odd question, so I've included the motivation below the fold.*


Throughout we work in ZFC+"$\omega\_1^r$ is countable for all $r\in\mathbb{R}$:"


Say that a set $X\subset\omega\_1$ is *really climbable* if $X$ is countable *(that's the silly case)* or there is some real $r$ and some $f\in L[r]$ such that for all countable $\alpha$, $$f(X\cap\alpha)\in X\setminus \alpha.$$ *(Note that we don't demand $f(X\cap \alpha)=\min(X\setminus\alpha)$.)* That is, an uncountable $X$ is really climbable if there is some real which lets us build a way to climb up $X$. Obviously sufficiently nice sets are really climbable, as are sets *containing* sufficiently nice sets; at the same time, under CH it's easy to build a non-really-climbable set. My question is:



> 
> What can be said about the really climbable sets without CH?
> 
> 
> 


In particular, it's not clear to me that "Every set is really climbable" is impossible.




---


Motivation: I'm trying to answer a question I left open in my thesis, which has been bothering me for some time *(which was also ultimately, but much directly, the motivation for [this other question](https://mathoverflow.net/questions/278780/conflating-reals-and-sets-of-countable-ordinals-nicely) of mine)*:


For $K\subseteq\omega\_1$, consider the game (on $\omega$) $Copy(K)$ where


* Player I builds a single linear order $A$
* Player II builds a list of linear orders $B\_i$ ($i\in\omega$)
* Player II wins iff $(i)$ **every** $B\_i$ is isomorphic to some $\alpha\_i\in K$, and $(ii)$ if $A\cong\beta$ with $\beta\in K$, then $\beta=\alpha\_i$ for some $i\in\omega$. *Here I conflate an ordinal $\gamma$ with the linear order $(\gamma, <)$.*


That is, player II is trying to "guess" what element of $K$ player I is playing; if player I can trick player II into building something not in $K$, they win, and otherwise they win if they build something not isomorphic to anything built by II. *(This is an instance of [a class of games defined by Montalban](https://math.berkeley.edu/~antonio/papers/CopyableStructures.pdf).)*


By $\Sigma^1\_1$-bounding, if $K$ is cofinal in $\omega\_1$ then $Copy(K)$ can't be a win for player II, and it's easy to see that $Copy(K)$ *is* a win for II if $K$ is countable. And assuming CH, there is some $K$ such that $Copy(K)$ is undetermined. However, if we drop CH, then I don't know how to produce an undetermined $Copy(K)$, even using Choice - roughly, the reason is that we have continuum-many strategies to defeat but only $\omega\_1$-many choices to make in building $K$ (and of course it's even consistent that there are as many strategies as there are subsets of $\omega\_1$). "Obviously" there is still such a $K$, but the question is (so far as I know) open for now.


The question above is one which grows naturally out of this problem: intuitively, if it's consistent that every set of countable ordinals is really climbable, then it would be more plausible that "$Copy(K)$ is determined for all $K\subseteq\omega\_1$" could be consistent with ZFC.



Climbing up subsets of $\omega_1$ using reals
"Every set is really climbable": this contradicts AC. By AC construct a set $X\subseteq \omega\_1$ such that $L[X]\models $"$\omega\_1 =\omega\_1^V$". Then for arbitrarily large $\gamma < \omega\_1$ we have that $L\_\gamma[X\cap \gamma] \models $"*Every set is countable*". But such a set cannot be climbable, for if $f\in L[r]$ were a 'climbing' function, according to the definition we should have such $X\cap \gamma \in dom(f) \subset L[r]$ contradicting the inaccessibility of $\omega^V\_1$ there. 


On the other hand, assuming $AD$, for example in $L({R})$ assuming large cardinals (where also DC holds - but that is not relevant), by an early result of Solovay, every subset of $\omega\_1$ is in some $L[r]$ for a real $r$; clearly then every such subset is really, and trivially, climbable.


