Let $A$ denote Ackermann set theory, as laid out e.g. here. It is a standard result due to Levy and Reinhardt that $A$ and $ZF$ are mutually interpretable in conservative extensions of one-another.

How do we see that the empty class is a set in $A$?

Despite scanning the relevant papers I seem unable to find a simple argument laying out how $\emptyset\in\mathbb{V}$. Naively, my thought was to consider the predicate $$\phi\equiv\ "x\ \text{is a set}\wedge x\neq x"$$ and take $\emptyset$ to be the set guaranteed by the reflection schema (axiom $3$ above) together with $\phi$, but I am concerned that by writing '$x$ is a set' in the definition of $\phi$ I am implicitly allowing $\mathbb{V}$ to appear, since this is equivalent to $x\in\mathbb{V}$ -- this would obviously prevent us from defining $\emptyset$ in this fashion (unless I am misunderstanding what it means for a constant symbol to 'be a parameter' in a predicate).

Does this approach work, or is some other trick required? Any tips are appreciated.


1 Answer 1


I believe you're overthinking the reflection scheme. Just use the formula $F(x)\equiv x\not=x$. Then:

  • We trivially have $\forall x(F(x)\rightarrow x\in V)$, since no $x$ at all satisfies $F$.

  • Consequently, reflection applies, and we get $\{x:F(x)\}\in V$ as desired.

  • $\begingroup$ Nice, I was literally on my way back to the computer to ask if this was one of those 'vacuous implication' things. Thanks as always Noah. $\endgroup$
    – Alec Rhea
    May 16 at 1:21

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