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In ZF(C), one can easily get a class partition of $V$, we can even get an $\mathrm{Ord}$-partition using the Cumulative hierarchy: $P=\{V_{α+1}\setminus V_α\mid α∈\mathrm{Ord}\}$, such a partition let us do stuff like Scott's trick: given a class $A$, we can look at $A∩V_β$ where $β$ is the minimal $β$ so that intersection is not empty.

But the fact that $\bigcup P=V$ is equivalent to the axiom of regularity.

I remember somewhere reading that in $ZF\text{-regularity}$ we can't have Scott's trick like trick.

We can formulate Scott's trick as:

There exists an $\mathrm{Ord}$-partition of $V$ (or equivalently - there exists cumulative hierarchy that sums up to the universe).

In a sense this version of Scott's trick is a "global trick".

While the intuition tells me that $\text{regularity}$ does not follow from this version of Scott's trick, is this true?

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    $\begingroup$ The “local Scott’s trick” as formulated here is trivially provable for every $\varphi$ (that is, ST holds): if $\{x:\varphi(x,x_1,\dots)\}=\emptyset$, put $S_\varphi(x_1,\dots)=\emptyset$, otherwise pick $x$ such that $\varphi(x,x_1,\dots)$ and put $S_\varphi(x_1,\dots)=\{x\}$. $\endgroup$
    – Emil Jeřábek
    Commented Aug 22, 2021 at 18:59
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    $\begingroup$ To see that GST is strictly weaker than regularity, GST holds if the universe equals the cumulative hierarchy over a set of Quine atoms. If the set of atoms is nonempty, the axiom of regularity fails. This is all consistent relative to ZF. $\endgroup$
    – Emil Jeřábek
    Commented Aug 22, 2021 at 19:05
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    $\begingroup$ You can take any model of ZFC and add a single Quine atom. This will not spoil global choice. $\endgroup$
    – James E Hanson
    Commented Aug 22, 2021 at 19:36
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    $\begingroup$ Your definition of $P$ seems inside-out. You want $P$ to be the set of difference sets, not the set of ordinals. $\endgroup$
    – Joel David Hamkins
    Commented Aug 22, 2021 at 21:07
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    $\begingroup$ @Holo: Between "don't read this part, it's trivial" and strike-through text, maybe it's worth just rewriting the question to be clear? If someone wants to see the original, they can still do that. $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 23, 2021 at 8:07

1 Answer 1

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Yes! Extend $\sf ZF - Reg.$ with the existence of a unique Quine atom $\sf Q=\{Q\}$, take $V$ to be the hierarchy over $\sf Q$, that is: $$\begin{align} & V_\emptyset = \sf Q \\ & V_{\alpha+1}= \mathcal P (V_{\alpha}) \\ & V_\lambda= \bigcup_{\alpha < \lambda} V_\alpha, \text {for limit } \lambda \\ & V= \bigcup _{\alpha \in \mathrm{Ord}} V_\alpha \end {align}$$

clearly this $V$ has an $\mathrm{Ord}$-partition.

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