Is there a difference between using $\alpha \cup \{\alpha\}$ and $\mathscr{P}(\alpha)$ in constructing ordinals in the cumulative hierarchy of sets?

Question

In constructing the sets in the cumulative hierarchy from the empty set, why is the power set operation even needed?

Let $V_{0} = \emptyset$, and inductively define $V_{\alpha+1} = V_{\alpha} \cup \{V_{\alpha}\}$. By using the union operator we can keep on constructing larger and larger sets. Why do we even need the power set operator here?

Yes, you can build *a* cumulative hierarchy this way (cf. https://en.wikipedia.org/wiki/Cumulative_hierarchy), but by changing the recipe it is no longer *the* cumulative hierarchy as introduced by von Neumann/Zermelo — 2017-02-16

user id:622 311k · Accepted Answer

This way you only get the ordinals. Note that the construction only yields sets which are transitive and all of their elements are transitive.

But not all sets are ordinals, not all sets are even sets of ordinals. Even if every set is equipotent to an ordinal (assuming choice, anyway).

So in some sense, the power set operator is more powerful? – 2017-02-16 — 2017-02-16
Well... Obviously. Just look at Cantor's theorem. – 2017-02-16 — 2017-02-16

Is there a difference between using $\alpha \cup \{\alpha\}$ and $\mathscr{P}(\alpha)$ in constructing ordinals in the cumulative hierarchy of sets?

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