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Let $\mathbb P$ be a forcing that does not collapse $\omega_1$, $\theta$ sufficiently large and regular and $X\prec H_\theta$ a countable elementary substructure with $\mathbb P\in X$ as well as $p\in \mathbb P\cap X$. This is a typical situation if one deals with (semi)proper forcings. Let $\delta=X\cap\omega_1$. Assume further that $\dot S\in X$ is a $\mathbb P$-name for a subset of $\omega_1$. I would like to find a $(X, \mathbb P)$-semigeneric condition $q\leq p$ and simultaneously control whether $\delta\in\dot S^G$ or not. Lets say we ask for such $q$ so that $$q\Vdash \check\delta\notin\dot S$$ There are obvious obstructions to this: If there is $T\in \mathcal P(\omega_1)\cap X$ with $\delta\in T$ and $$p\Vdash\check T\subseteq \dot S \mod \mathrm{NS}_{\omega_1}$$ then there cannot be such $q$. I can prove that if the Strong Reflection Principle ($\mathrm{SRP}$) holds, this is the "only reason" for the nonexistence of such $q$:

Assume $\mathrm{SRP}$ holds. Let $\mathbb P, X, p, \dot S$ as above. Exactly one of the following holds:

  1. There is $q\leq p$ that is $(\mathbb P, X)$-semigeneric and $q\Vdash\check\delta\notin \dot S$
  2. There is $T\in \mathcal P(\omega_1)\cap X$ with $\delta\in T$ and $$p\Vdash\check T\subseteq \dot S \mod \mathrm{NS}_{\omega_1}$$

The proof of $\neg 1.\Rightarrow 2.$ is similar to the the proof of "stationary set preserving forcings are semiproper" (under $\mathrm{SRP}$). Let us say the name dichotomy holds for $\mathbb P$ if exactly one of 1. and 2. holds for any $X, p, \dot S$ as above. One can hope that if a property is provable for the class of $\omega_1$-preserving forcings under $\mathrm{SRP}$ then maybe this property holds in general for more "well-behaved" classes of forcings. So my question is:

Is it provable (in $\mathrm{ZFC}$) that the name dichotomy holds for

  • $\sigma$-closed forcings?
  • proper forcings?
  • semiproper forcings?
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  • $\begingroup$ I probably should have added that the name dichotomy is easily provable for c.c.c. forcings. $\endgroup$ Jul 5, 2022 at 16:03
  • $\begingroup$ Do you know a class of forcing for which it definitely does not hold? $\endgroup$ Jul 8, 2022 at 13:06
  • $\begingroup$ Stationary set preserving forcings that are not semiproper, like Namba forcing in $L$, are counterexamples. I do not know any other unfortunately. $\endgroup$ Jul 8, 2022 at 15:03

1 Answer 1

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It turns out that it is not provable in $\mathrm{ZFC}$ that even $\sigma$-closed forcings satisfy the name dichotomy. The answer to all three questions is thus no (unfortunately).

I will show that under $V=L$, $\mathbb P:=\mathrm{Add}(\omega_1, 1)$ does not satisfy the name dichotomy:

Consider the function $f:\omega_1\rightarrow \omega_1$ defined via $f(\alpha)$ is the least $\beta$ so that $\alpha$ is countable in $L_{\beta+1}$ (this is the wellknown example of a function in $L$ not bounded by any canonical function). If $G$ is $\mathbb P$-generic, then in $V$ we can define the following set: $$S=\{\alpha<\omega_1\mid G\upharpoonright\alpha\text{ is generic over }L_{f(\alpha)}\}$$ Here, "$G\upharpoonright \alpha$ is generic over $L_{f(\alpha)}$" means that $G\upharpoonright \alpha=\{p\in G\mid \mathrm{dom}(p)\subseteq\alpha\}$ is generic for $\mathrm{Add}(\alpha, 1)^{L_{f(\alpha)}}$ over $L_{f(\alpha)}$ (in particular $G\upharpoonright\alpha\subseteq L_{f(\alpha)}$). Let $\dot S$ be a name in $V$ for this set $S$.

Claim: $S$ is stationary costationary in $V[G]$.

Proof: It is not difficult to see that $S$ is stationary. To see that $S$ is costationary, let $\dot C$ be a name for a club in $V$. Let $\theta$ be large enough, regular and $X\prec H_\theta$ countable with $\dot C\in X$. There is a descending sequence $\vec p:=\langle p_\alpha\mid \alpha<\omega_1\rangle\in H_\theta$ of conditions in $\mathbb P$ so that for all $\alpha$ there is $\alpha\leq\beta$ so that $p_\alpha\Vdash\check\beta\in \dot C$. By elementarity, we may assume $\vec p\in X$. Let $q=\bigcup_{\alpha<\delta^X} p_\alpha$. It is clear that $q$ is not $(X,\mathbb P)$-generic, but we have $q\Vdash\check \delta^X\in \dot C$. It follows that $q\Vdash\check\delta^X\in\dot C-\dot S$. $\square$

The argument above shows even a little more: If $T\in V$ is stationary in $\omega_1$, then $T\cap S$, $T-S$ are stationary in $V[G]$ (just choose $X$ with $\delta^X\in T$). It follows that 2. of the name dichotomy fails for $\dot S$ (for any appropriate $\theta$, $X$, $p$).

To get a failure of 1. as well, it is enough to find a countable $X\prec H_\theta$ so that if $X\cong L_\gamma$ then $\mathcal P(\delta)\cap L_\gamma=\mathcal P(\delta)\cap L_{f(\delta)}$ for $\delta=\delta^X$: Any $q$ that is $(X, \mathbb P)$-semigenric is actually $(X, \mathbb P)$-generic (as $\mathbb P$ has size $\omega_1$) and hence as $L_\gamma$ and $L_{f(\delta)}$ have the same dense subsets of $\mathrm{Add}(\delta, 1)$, such $q$ forces $G\upharpoonright \delta$ to be generic over $L_{f(\delta)}$.

The following example of such an $X$ is due to Ralf Schindler:

Let $X_0=\mathrm{Hull}^{H_\theta}(\emptyset)$ and for $n<\omega$ let $X_{n+1}=\mathrm{Hull}^{H_\theta}(\{X_n\})$. Put $X=\bigcup_{n<\omega} X_n$. I claim that this $X$ works. Let $\delta=\delta^X$. Let $X\cong L_\gamma$.

Claim: $f(\delta)=\gamma+1$.

Proof: $L_\gamma$ is a model of $\mathrm{ZF}^-$, $\delta$ is uncountable in $L_\gamma$ and so $f(\delta)>\gamma$. For $n<\omega$ let $X_n\cong L_{\gamma_n}$. Then $\langle \gamma_n\mid n<\omega\rangle$ is definable over $L_{\gamma+1}$: $\gamma_0$ is the least $\xi$ with $L_\xi\prec L_{\gamma}$ and $\gamma_{n+1}$ is the least $\gamma_n<\xi$ with $L_\xi\prec L_{\gamma}$. Thus $\langle \delta^{X_n}\mid n<\omega\rangle$ is definable over $L_{\gamma+1}$ and as $\delta=\mathrm{sup}_{n<\omega} \delta^{X_n}$, $\delta$ is countable in $L_{\gamma+2}$. $\square$

As $L_\gamma$ is a model of $\mathrm{ZF}^{-}$ we have $$\mathcal P(\delta)\cap L_\gamma=\mathcal P(\delta)\cap L_{\gamma+1}$$ so that $X$ has the desired property.

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