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".




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



Scott's trick without regularity
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.


