Consider the following statement: (T) "If $M$ is a standard class model of ZFC isomorphic to $V$, then $M = V$." The statement (T) is equivalent to: "If the transitive collapse of a standard class model $M$ of ZFC is equal to $V$, then $M = V$." This is because the transitive collapse of a class $M$ is the unique transitive class that is elementhood-wise isomorphic to $M$.

Here, by standard class model of ZFC I mean a class model of ZFC whose elementhood relation is the real elementhood relation.

Assume that ZFC is consistent. Does ZFC prove (T)? Does ZFC disprove (T)? If no to both, does ZFC with some additional large cardinal axiom disprove (T)?

  • 1
    $\begingroup$ Note that $(T)$ can't actually be formulated as a single sentence in the language of set theory - it has to be formulated as a scheme (or we have to work in $\mathsf{NBG}$ or similar). $\endgroup$ Aug 19, 2020 at 22:20

1 Answer 1


No. Define $F:V\to V$ by $\in$-recursion as $F(x)=\{F(y):y\in x\}\cup\{\emptyset\}$. Clearly $F(x)$ is nonempty for all $x$. Also, $F$ is injective: if $F(x)=F(x')$, then by induction on $\max(\operatorname{rank}(x),\operatorname{rank}(x'))$ we may assume $F$ is injective on $x\cup x'$. Since $F(x)=F(x')$ we must have $\{F(y):y\in x\}=\{F(y):y\in x'\}$, but since $F$ is injective on $x\cup x'$ this implies $x$ and $x'$ have the same elements and thus $x=x'$. Also clearly $y\in x$ implies $F(y)\in F(x)$, and the converse follows from injectivity of $F$.

Taken all together, this shows that $F$ is an isomorphism from $(V,\in)$ to $(M,\in)$ where $M$ is the image of $F$. But $M\neq V$, since $\emptyset\not\in M$.

  • $\begingroup$ Sorry, there's one apparent consequence of this that really confuses me, that maybe you can explain. Let πœ™(π‘₯) denote the statement that π‘₯ lies in the image of your function 𝐹, so that 𝑀={π‘₯:πœ™(π‘₯)}. Then ZFC proves that βˆ…βˆ‰π‘€, hence that ¬βˆ€π‘₯πœ™(π‘₯). However, obviously ZFC does not prove this statement relativized to your model 𝑀, since that would be the contradictory statement ¬βˆ€π‘₯(π‘₯βˆˆπ‘€β†’πœ™(π‘₯)). So there's a theorem of ZFC that is refutable when relativized to 𝑀. What is wrong with this argument? Am I confusing ZFC with the metatheory somehow? $\endgroup$ Aug 20, 2020 at 5:59
  • 1
    $\begingroup$ ZFC does prove it relativized to $M$. The relativization is not just $\neg\forall x(x\in M\to\phi(x)$ but rather $\neg\forall x(x\in M\to\phi^M(x)$ where $\phi^M$ is the relativized version of $\phi$. Since $\phi^M$ and $\phi$ are not the same this is not a contradiction. $\endgroup$ Aug 20, 2020 at 14:41
  • $\begingroup$ Gotcha, and thank you. Can you by any chance describe the class corresponding to the relativization $\phi^M$? Is that just $F(F(V))$? Which classes $M$ have the property (like $L$) that they equal their relativization, or at least that $\forall x (\phi(x) \rightarrow \phi^M(x))$? (Restrict attention to models of ZF if you like.). Sorry for the follow-up questions. $\endgroup$ Aug 21, 2020 at 19:38
  • 1
    $\begingroup$ I don't think there's any simple concrete description of $\phi^M$; it's just what you get by doing the same construction inside $M$. It's not $F(F(V))$ but $F^M(F(V))$ where $F^M$ is $F$ with its definition interpreted inside $M$. I don't know much you can say about the latter question; certainly most classes don't have that property. $\endgroup$ Aug 21, 2020 at 19:54
  • $\begingroup$ I was thinking the property might relate to the transitive classes somehow. I'm trying to figure out what's special about $L$. Do you know any other classes besides $L$ that have that property? One thing is clear I think is that if $M = \{x: \phi(x)\}$ is such a class model of ZF, then ZF cannot refute $V = M$. $\endgroup$ Aug 21, 2020 at 19:59

Your Answer

By clicking β€œPost Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.