# Can we force the existence of this function?

I want to see if it is possible to force the existence of a function $F:\aleph_2 \times \aleph_2\rightarrow \aleph_1$ such that:

a) $F(a,b)=F(b,a)$, for all $a,b\in \aleph_2$ and

b) for all distinct $a,b$, the set $\{x|F(a,x)=F(b,x)\}$ is finite.

What is known:

1) Under CH, there is no such function.

2) If such a function exists, let A be a subset of $\aleph_2$ of size $\aleph_1$. Consider the functions $F(a,\cdot)$ restricted to A. We have a family of $\aleph_2$ many such functions from $\aleph_1$ to $\aleph_1$ and any two of them agree only on a finite set. That's a strongly almost disjoint family of size $\aleph_2$. Baumgartner call this $A(\aleph_1,\aleph_2,\aleph_1,\aleph_0)$. It is consistent together with the negation of CH that either $A(\aleph_1,\aleph_2,\aleph_1,\aleph_0)$ or its negation hold. In particular, $\neg CH$ + "there is no such function" is consistent.

I want to see if anyone knows any result on the positive side.

• Do you know the answer without condition a? – Will Sawin Jan 2 '14 at 19:44
• @WillSawin: No. – Ioannis Souldatos Jan 2 '14 at 20:13
• @WillSawin: It is not very different. If (a) is missing then (b) must include that the sets {x|F(x,a)=F(x,b)} are finite. – Ioannis Souldatos Jan 2 '14 at 20:28
• Here's an idea: the forcing conditions consist of a finite symmetric partial function $f$ from $\aleph_2\times \aleph_2$ to $\aleph_1$ and a finite subset $A$ of $\aleph_2$. A condition $(f',A')$ is stronger than $(f,A)$ if $f'$ and $A'$ extend $f$ and $A$, respectively, and if for any (coded) pair $\langle a,b\rangle\in A$ any new values that $f'$ assigns to the same rows of the $a$-th and $b$-th columns are distinct. Taking as $F$ the union of the first components of a generic filter, it seems to me that this satisfies both of your requirements. – Miha Habič Jan 3 '14 at 20:26
• Miha: The f-values taken on $\omega$ of the conditions in the generic set will be cofinal in $\omega_1$, therefore collapsing $\aleph_1$. – Péter Komjáth Jan 4 '14 at 7:01

First notice that we can separate the condition depending the relative position of $a,b$, and $x$ in (b): let $F_0$ be such that for any $a,b$ there are finitely many $x<\min(a,b)$ with $F_0(x,a)=F_0(x,b)$, let $F_1$ be likewise for $a<x<b$ and let $F_2$ be for $a,b<x$. Then $F(a,b)=\langle F_0(a,b),F_1(a,b),F_2(a,b)\rangle$ is as required. $F_2$ obviously exists as we can have $F_2(a,x)\neq F_2(b,x)$ for $a\neq b<x$. It is well known that there is $F_1:[c]^2\to\omega$ with $F_2(a,x)\neq F_2(x,b)$ for $a<x<b$ ($c$ is continuum). So there remains the task to force $F_0$ and this can be done the way Miha suggested.

Let $F:[\omega_2]^2\to [\omega_2]^{\leq\aleph_0}$ be a Delta-function. We force with conditions $(s,f)$ where $s\in[\omega_2]^{<\omega}$, $f:[\omega_2]^2\to\omega$, if $x<a,b$ are in $s$, $f(x,a)=f(x,b)$, then $x\in F(a,b)$, and $(s',f')\leq (s,f)$ if $s'\supseteq s$, $f'\supseteq f$, and if $a,b\in s$, $x\in s'-s$, $x<a,b$, then $f'(x,a)\neq f'(x,b)$.

The essential thing is to show ccc. Assume that we have $p_\alpha=(s\cup s_\alpha,f_\alpha)\in P$, and $f_\alpha|s$ are the same. By removing countably many $p_\alpha$, we can get thet $s_\alpha\cap F(a,b)=\emptyset$ for $a,b\in s$. By the Delta-function property, there are $\alpha<\beta$ such that if $x<a,b$, $x\in s$, $a\in s_\alpha$, $b\in s_\beta$, then $x\in F(a,b)$. Now we can extend $p_\alpha$ and $p_\beta$ to a common $p=(s',f')$ such that $f'(u,v)$ ($u\in s_\alpha$, $v\in s_\beta$) differ from each other and the range of $f_\alpha, f_\beta$.

The consistency of the existence of a Delta function is in James E. Baumgartner and Saharon Shelah, Remarks on superatomic Boolean algebras, Ann. Pure Appl. Logic 33 (1987), no. 2, 109–129.

• @Komjath: Your answer looks very promising. Give me a couple of days before I look at it, because we just returned from travelling and I am trying to get into the routine again. – Ioannis Souldatos Jan 6 '14 at 21:11

While this is extremely late, I previously ran into something similar when looking for examples of objects on $\omega_2$ that can be forced with side conditions, and I think the following method has certain advantages. If you want to avoid the use of a Delta-function and force directly, using the poset that you'd imagine, this can be done and is one of the easiest applications of Neeman's method of forcing with sequences of models of two types." The initial forcing one might try is to let $\mathbb{P}$ consist of $p=(f_p,A_p)$, where $f_p : \omega_2 \times \omega_2 \rightarrow \omega_1$ is a finite partial symmetric function and $A_p$ is the set of first coordinates in the domain of $f_p$, where $(f_q, A_q) \leq (f_p,A_p)$ if $f_p \subseteq f_q$ and for every $\delta \in A_q \setminus A_p$ and $\alpha \neq \beta \in A_p$, $f_q(\alpha,\delta) \neq f_q(\beta,\delta)$. However, as noted by Péter Komjáth, $F_G '' \omega \times \omega$ will be cofinal in $\omega_1$. Furthermore, because we're also worried about collapsing $\omega_2$, only considering conditions $p = (f_p, A_p, \mathcal{M}_p)$, where $\mathcal{M}_p$ is a finite $\in$-chain of countable elementary substructures of $H_{\omega_2}$, with the interaction requirement that if $\alpha, \beta \in A_p \cap M$, then $f_p(\alpha,\beta) \in M$ for every $M \in \mathcal{M}_p$, will result in a proper poset, but will not work because we cover $H_{\omega_2}$ by an increasing chain of countable structures, so in particular $\omega_2$ is collapsed.

However, if we let $\mathcal{M}_p$ instead be a finite $\in$-chain of elementary submodels of $H_{\omega_2}$ that can be either countable or internally approachable (for example) closed under intersections, with the same interaction requirement (which says nothing for the larger models), then it's not too difficult to show that $\mathbb{P}$ is strongly proper for the set of countable elementary submodels of $H_{\omega_2}$ and for the set of internally approachable elementary submodels of $H_{\omega_2}$, and because the former set is club and the latter is stationary, in $[H_{\omega_2}]^{\omega}$ and $[H_{\omega_2}]^{\omega_1}$, respectively, $\omega_1$ and $\omega_2$ are preserved. By strongly proper here we mean that conditions within models can be extended to strongly generic conditions, where $p$ is $(M, \mathbb{P})$-strongly generic if for every $r \leq p$, there exists $r \restriction M \in M$ such that $q \leq r \restriction M$ with $q \in M$ implies that $q \parallel r$. There are a few details to check, but once you see a few examples of this method, this particular forcing is one of the easiest to consider.