If $\tau$ is a name, is $\{\sigma:\Vdash \sigma=\tau\}$ a proper class?

Question

Fix some (atomless) poset $\mathbb P$. Let $V^\mathbb{P}$ be the class of $\mathbb P$-names and $\tau\in V^\mathbb{P}$. Consider the collection $$ \{\sigma\in V^\mathbb{P} : \Vdash_\mathbb{P} \sigma=\tau\} $$ My question is: Can this collection be a proper class? I've always believed the answer to be "yes", but I'm struggling to actually prove this.

If $\tau=\emptyset$, then obviously the only name which is forced to be equal to $\tau$ is $\tau$ itself. Indeed, suppose $\sigma\neq \emptyset$, say $(\vartheta,p)\in \sigma$ for some $p\in\mathbb{P}$. Then $p\Vdash \vartheta \in \sigma$, so $p\Vdash \sigma\neq \tau$ and thus $\not\Vdash \sigma=\tau$.

Also, the above is false if we're taking the Boolean algebra approach to forcing. Indeed, since we have access to the bottom element $0$ saying false, we can consider, for any $x\in V$, $$ \sigma_x=\{(\check x,0)\}\cup \tau $$ Then $\{\sigma_x:x\in V\}$ is a proper class, and $\|\sigma_x=\tau\|=1$ for every $x$.

It is also false if we replace $\Vdash$ by $p\Vdash$ for some $p\in\mathbb{P}$. Indeed, if $p,q\in \mathbb{P}$ and $p\perp q$, put $\sigma_x=\{(\check x, q)\}$ for $x\in V$. Then $p\Vdash \sigma_x=\emptyset$, and the $\sigma_x$ are pairwise distinct.

I've thought about arguing that $$ \forall \tau\in V^\mathbb{P} \exists X_\tau\forall \sigma\in V^\mathbb{P} [\sigma \in X_\tau\iff \Vdash \sigma=\tau] $$ by $\in$-induction. Say $\tau\in V^\mathbb{P}$ and the statement holds for every $\sigma$ of smaller rank. Put $X=\bigcup\{X_\sigma:\sigma\in\text{dom}(\tau)\}$. Suppose there is a proper class of names which are forced to equal $\tau$. Pick a name $\vartheta$ with $\text{dom}(\vartheta)\setminus X\neq \emptyset$ and $\Vdash \vartheta=\tau$. There is some pair $(\pi,p)\in \tau$ with $\pi\not\in X$. We know that $\not \Vdash \pi=\sigma$ for every $\sigma\in \text{dom}(\tau)$, but this is not enough to derive a contradiction, because we only have $p\Vdash \sigma \in \tau$.

Answer 1

Your class is a proper class if $\mathbb{P}$ is atomless. Let $p$ and $\sigma$ be such that $\{\mu: p\Vdash \mu=\sigma\}$ is a proper class. Now consider $\rho=\{(\sigma,p)\}$. If $p\Vdash \mu=\sigma$ and $\rho':=\{(\mu,p)\}$, then

1. $p\Vdash \rho=\rho'$ (In fact, $p$ thinks $\rho$ is a singleton whose element is $\sigma$, and
2. If $q$ is incompatible with $p$, then $q\Vdash \rho=\rho'=\varnothing$.

Hence $\{r:r\Vdash\rho=\rho'\}$ contains a dense subset, namely $$\{r: r\le p \lor r\perp p\}.$$ Thus $1\Vdash \rho=\rho'$.