A model without a choice function

Question

Suppose we force over $\operatorname{Add}(\omega, \omega)$ with a $V$-generic $G$. Let $g = \bigcup G$ and $r_i$ be the $i$th real added, i.e $r_i = \{n : g(i, n) = 1\}$.

Then define (in $V[G]$) for all $i$, $R_i = \{r \subseteq \omega : |r \Delta r_i| < \aleph_0 \}$. Let $S = \{\{R_{2i}, R_{2i+1}\} : i < \omega\}$. Then consider the model $N = \mathrm{HOD}(\operatorname{trcl}\{S\})^{V[G]}$. I want to show that $N$ has no choice function for $S$.

I know that there is a choice function for $\{\{r_{2i}, r_{2i+1}\}:i < \omega\}$ in $N$ at least, and I have tried some homogeneity arguments. Unfortunately I haven't been able to get far. I would appreciate hints or help.

Answer 1

Let $p$ be a condition such that $p\Vdash\varphi(\dot X,\dot y,\check\alpha,\vec S_i)\text{ defines a choice function}$, i.e. $p$ forces that there is an ordinal-definable choice function with parameters in the transitive closure $S$.

Since $p$ is a finite condition in the Cohen forcing, and $\vec S$ is finite, we can assume for simplicity that if $(n,m)$ is in the domain of $p$, then $\{R_n,R_{n\pm1}\}$ or $R_n$ itself is in $\vec S$; and vice versa, if $R_n$ is in $\vec S$, then $(n,m)\in\operatorname{dom}(p)$ for some $m$, etc.. Here $\{R_n,R_{n\pm1}\}$ is the unique pair that contains $R_n$.

Pick some large enough $n$ such that $(n,m)$ is not in the domain of $p$, and therefore also not mentioned in $\vec S$ in the sense above. Let $q$ be a stronger condition such that without loss of generality, $q\Vdash\varphi(\{\dot R_n,\dot R_{n\pm1}\},\dot R_n,\check\alpha,\vec S)$, of course we can take $\dot R_{n\pm1}$.

Note that neither $n$ nor $n\pm1$ can appear in the original $p$ or $\vec S$. Let $m$ be large enough such that $(n,k)$ or $(n\pm1,k)\in\operatorname{dom}p$ implies $k<m$. Now consider the following automorphism $\pi$:

1. It acts on the left coordinate as the cycle $(n\ \ n\pm1)$,
2. On these two coordinates, $(n,k)$ is switched with $(n,m+k)$ for all $k<m$ (and likewise for $(n\pm1,k)$ that is).

In other words, $\pi$ "flips" between the $n$th and $n\pm1$th reals, and then on these two coordinates it also switches the first $m$ points with the next $m$ points to make sure that we end up in "a disjoint part of the domain".

We can show that:

1. $\pi p=p$ and $\pi q$ is compatible with $q$: the former is easy, since $n$ and $n\pm1$ are not mentioned in the domain of $p$; the latter is straightforward, we only changed two coordinates and it is easy to see that these are compatible since we also moved them "internally" to make them disjoint.

2. If $\dot x$ is in $\vec S$, then $\pi\dot x=\dot x$: this is also a straightforward consequence of $\pi p=p$ and the condition with which we selected $p$ and $\vec S$.

3. $\pi\dot R_n=\dot R_{n\pm 1}$ and $\pi\dot R_{n\pm1}=\dot R_n$.

Now we get that $\pi q\Vdash\varphi(\{\pi\dot R_n,\pi\dot R_{n\pm1}\},\pi\dot x,\pi\check\alpha,\pi\vec S)$. But the above tells us that this is the same as writing $\pi q\Vdash\varphi(\{\dot R_n,\dot R_{n\pm1}\},\pi\dot x,\check\alpha,\vec S)$.

Using now that $q$ and $\pi q$ are compatible that means that $$\pi q\cup q\Vdash\varphi(\{\dot R_n,\dot R_{n\pm1}\},\dot x,\check\alpha,\vec S)\land\varphi(\{\dot R_n,\dot R_{n\pm1}\},\pi\dot x,\check\alpha,\vec S),$$ but it is easy to show that $\varnothing\Vdash\dot x\neq\pi\dot x$, since $\dot x$ is either $\dot R_n$ or $\dot R_{n\pm1}$. And so this is a contradiction.