# Diagonalizing against $\omega_1$-sequences of functions mod finite

The following statement is a direct consequence of the Continuum Hypothesis:

There exists a sequence $$\langle f_\alpha:\omega_1\rightarrow\omega_1 ~ \vert ~ \alpha<\omega_1\rangle$$ of functions such that there is no function $$f:\omega_1\rightarrow\omega_1$$ with the property that the sets $$\{\xi<\omega_1 ~ \vert ~ f(\xi)=f_\alpha(\xi)\}$$ are finite for all $$\alpha<\omega_1$$.

Moreover, since a failure of this statement can be used to obtain an $$\omega_2$$-sequence of subsets of $$\omega_1$$ with pairwise finite intersection, results of Baumgartner in

Baumgartner, James E., Almost-disjoint sets, the dense set problem and the partition calculus, Ann. Math. Logic 9, 401-439 (1976). ZBL0339.04003.

show that the statement is not equivalent to CH.

Question: Can the above statement consistently fail?

• I think you have already looked at Laver’s paper: link.springer.com/content/pdf/10.1023/A:1004392507990.pdf Jun 21, 2021 at 20:39
• @ Rahman. M: Yes, I looked at Laver's paper. But, I do not see how to use his results to derive a failure of the statement if a given sequence satisfies $f_\alpha(\beta)\geq\alpha$ for all $\alpha,\beta<\omega_1$. Jun 22, 2021 at 14:01
• Can't you cook some proper/semi-proper/SSP poset that would have added such a sequence, and then apply PFA/SPFA/MM as appropriate? (Easier said than done, I know.) Jun 22, 2021 at 14:52
• Are you requiring ZFC? Jun 22, 2021 at 16:22
• (Just as a tip, if you put a space between @ and the username, the user won't be notified.) Jun 23, 2021 at 8:55

The arguments in Section 6 of my paper "The nonstationary ideal in the $$\mathbb{P}_{\mathrm{max}}$$ extension" show that there is a proper forcing adding a function from $$\omega_{1}$$ to $$\omega_{1}$$ which agrees with each such ground model function in only finitely many places. I say there that Todorcevic had done something similar in his "A note on the Proper Forcing Axiom". It follows that under PFA, and in the $$\mathbb{P}_{\mathrm{max}}$$ extension, the statement above fails. You don't need any large cardinals, however.

• Great! This is the result I was looking for. Thanks a lot! Jul 13, 2021 at 9:31

This isn't an answer, as you're working in ZFC. But it seems worth noting. Assume ZFC + AD$$^{L(\mathbb{R})}$$. Then $$L(\mathbb{R})$$ satisfies ZF + AD + DC + "the statement is false".

Proof: Work in $$L(\mathbb{R})$$. Then every real has a sharp, and every subset of $$\omega_1$$ is constructible from a real. Suppose $$\vec{f}=\left_{\alpha<\omega_1}$$ witnesses Statement. Let $$x\in\mathbb{R}$$ be such that $$\vec{f}\in L[x]$$. Define $$g:\omega_1\to\omega_1$$ by $$g(\alpha)=$$ the $$(\alpha+1)$$th $$x$$-indiscernible. So $$g$$ is injective, strictly increasing, and its range consists of $$x$$-indiscernibles. Suppose $$\beta<\omega_1$$ and $$A\subseteq\omega_1$$ is infinite and $$f_\beta\upharpoonright A=g\upharpoonright A$$. Fix a finite set $$s$$ of $$x$$-indiscernibles such that $$f_\beta$$ is definable in $$L[x]$$ from $$(s,x)$$. Since $$gA$$ is infinite, we can choose $$\iota\in(gA)\backslash s$$. Let $$\xi=g^{-1}(\iota)$$. Since $$f_\beta(\xi)=\iota$$, $$\iota$$ is definable over $$L[x]$$ from $$(s,\xi,x)$$. But $$\xi<\iota$$, so we can choose a finite set $$t$$ of $$x$$-indiscernibles such that $$\xi$$ is definable from $$(t,x)$$, with $$\iota\notin t$$. So $$\iota$$ is definable over $$L[x]$$ from $$(s\cup t,x)$$, where $$s\cup t$$ are $$x$$-indiscernibles and $$\iota\notin s\cup t$$, and this contradicts that $$\iota$$ is an $$x$$-indiscernible.

Remark: The $$\mathbb{P}_{\mathrm{max}}$$ extension of $$L(\mathbb{R})$$ inherits some related properties, and it seems tempting to try to do a version of the preceding argument there, but I haven't seen how to do that.

• Where are you using $\mathrm{AD}$? It seems like your argument would work under the assumption that every real has a sharp and $V=L(\mathbb R)$. Jun 24, 2021 at 9:23
• @YairHayut Also to get every subset of $\omega_1$ is constructible from a real (which, together with the sharps for reals, contradicts AC, as it implies that the club filter on $\omega_1$ is an ultrafilter). (But ZF + "every real has a sharp" + "every subset of $\omega_1$ is constructible from a real" is enough.) Jun 24, 2021 at 9:44
• If I understand $\Bbb P_{\max}$ correctly (which is a very questionable assumption), it essentially adds a sequence of length $\omega_2$ to the universe. It stands to reason, then, that any subset $\omega_1$ only codes for a small intermediate model, and that there is something to be said about the quotient forcing there. If that is indeed the case, then it might be enough to show that there is enough genericity between subsets of $\omega_1$ to preserve the failure of "the statement" (since any sequence of functions is just a subset of $\omega_1$ in disguise). Jun 24, 2021 at 14:06
• @Farmer S: Thanks a lot for your post! Jun 24, 2021 at 21:26
• @Asaf Karagila♦: Theorem 7.7 in Larson's handbook article shows that the $\mathbb{P}_{max}$-extension of $L(\mathbb{R})$ is "minimal", in the sense that every subset of $\omega_1$ added by $\mathbb{P}_{max}$ already generates the entire extension. Jun 24, 2021 at 21:34