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 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.)

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.


2 Answers 2


"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.


Concerning Copy(K)-determinacy: you mention that every set of countable ordinals being climbable would lend plausibility to this in ZFC (that is with AC). We just saw that climbability fails under AC; of course $\forall K$Copy(K)-determinacy holds under AD; and if you make the game harder for II by insisting (I make it that II has the winning strategy, not I, by Boundedness) that II's wellorders code all of $K$ below their supremum, then one can conclude that $K$ is in $L[\tau]$ where $\tau$ is II's strategy. So for this slightly more rigorous game, its determinacy for all $K$ is inconsistent with choice and $\omega_1$ inaccessible to reals.

  • $\begingroup$ Sorry for the late response; I don't see why II has the winning strategy. It seems to me that I still has a winning strategy, if it exists, by bounding. $\endgroup$ Jan 29, 2018 at 15:22

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