TITLE: Scott's trick without regularity
QUESTION [11 upvotes]: 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?

REPLY [4 votes]: 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.