# Finding a winning strategy for a game in ZF + “$\omega_1$ is regular”.

Say two players ($$A$$ and $$B$$) are playing a game of length $$\omega$$ on $$\omega_1$$ in which they play a countable ordinal turn by turn (starting with $$A$$) and form a sequence of countable ordinals, $$(a_0,b_0,a_1,b_1, \ldots)$$ (they can see each other’s moves) of length $$\omega$$. Player $$B$$ wins if the set of the (values of the) obtained sequence is an ordinal, $$A$$ wins otherwise. Can player $$B$$ have a winning strategy in ZF + “$$\omega_1$$ is regular”(by which I mean, “Can ZF + ”$$\omega_1$$ is regular” prove that there is a winning strategy for $$B$$?”)?

Obvious strategy is to play every ordinal under every ordinal played by the opponent, which requires choice of some form, and I am not seeing how to do this using only the fact that $$\omega_1$$ is regular.

Note that converse is trivially true so if the answer is “yes”, then we have an equivalence. Thank you.

• @NoahSchweber Oops I totally forgot the most important part... I think choice is needed to choose enumerations of countable ordinals.
– Tan
Nov 12, 2021 at 20:47
• @AsafKaragila Sure: via choice fix a sequence $(f_\alpha)_{\alpha<\omega_1}$ of surjections $f_\alpha: \omega\rightarrow \alpha$, and on move $\langle m,n\rangle$ have player $B$ play $f_{\alpha_m}(n)$. (I'm assuming the OP is asking whether $\mathsf{ZF}$ + "$\omega_1$ is regular" proves that $B$ has a winning strategy; I guess you could also read it the other way, and then your comment is a hint to the OP. FWIW I think that the answer to my read of the question is negative.) Nov 12, 2021 at 20:51
• Doesn't a winning strategy for B obviously give a way to choose a bijection from every countably ordinal to $\omega$ uniformily? This will fail in the Solovay model while DC holds (and an innaccessible is clearly necessary). Nov 12, 2021 at 22:36
• @aschepler "regular" here is in the sense of "regular cardinal": the cofinality of $\omega_1$ is $\omega_1$. Nov 13, 2021 at 0:49
• @AsafKaragila Aren’t those already the moves of player $A$?
– Tan
Nov 13, 2021 at 8:58

From Noah's comment, we see that if there is a sequence $$\langle e_\alpha : \alpha < \omega_1 \rangle$$, where each $$e_\alpha \colon \omega \to \alpha$$ is a surjection, then Player B has a winning strategy.
On the other hand this is easily seen to be necessary for a winning strategy as well. Namely if $$\sigma$$ is a winning strategy for B and $$\alpha < \omega_1$$ is given, we play a game according to $$\sigma$$ where A constantly plays $$\alpha$$. As $$\sigma$$ is winning, the plays of $$B$$ enumerate some ordinal greater or equal than $$\alpha$$ in type $$\omega$$.
I claim that "$$\omega_1$$ is regular" is in general not sufficient. The most basic example is the so called Solovay model. Namely, in this model DC (dependent choice) holds, so $$\omega_1$$ is regular. But there is no injection from $$\omega_1$$ into the reals. A sequence $$\langle e_\alpha : \alpha < \omega_1 \rangle$$ as above on the other hand does give an injection from $$\omega_1$$ by associating each $$e_\alpha$$ to a well-order on $$\omega$$ of type $$\alpha$$ (which is a subset of $$\omega\times\omega$$, and thus a real).
The Solovay model uses an inaccessible cardinal in $$L$$ and in fact this is necessary here. If $$\omega_1$$ is regular, then it is also regular in $$L$$. If it was a successor, say $$\kappa^+$$, then, since choice holds in $$L$$, we have a sequence of surjections $$f_\alpha \colon \kappa \to \alpha$$ for every $$\alpha < \kappa^+$$. Since $$\kappa < \omega_1$$ in $$V$$, we have a surjection $$f \colon \omega \to \kappa$$ in $$V$$. But now we can combine these functions to get our $$e_\alpha$$'s.