# Is it consistent with Z - Regularity to have a set that is bigger than any set in the cumulative hierarchy of Z?

Edit: the question was answered to the negative because $ZF$ proves the existence of Hartog numbers. So this calls for a modification of the question to be in just $Z-\text{Regularity}$

Is it consistent with $Z-\text{Regularity}$ [instead of $ZF-\text{Regularity}$ in the original question] to have a set that is strictly bigger in cardinality than any set in the cumulative hierarchy of $Z$?

Formally: Is it consistent to have $Z-\text{Regularity}$ plus

$\exists x\, \forall y\, \big[\exists \alpha\, (y \in V_{\alpha}) \to \exists f\, \left(f\colon y \to x \wedge\, f \text{ is injective}\right) \wedge \not \exists g\, (g\colon x \to y \wedge\, g \text{ is injective})\big]$

Where $V_\alpha$ is defined in the usual manner.

The version of $Z$ defined here have all stages of $ZF$ below $V_{\omega+{\omega}}$

• Wait, even without Choice, every set is itself somewhere in the cumulative hierarchy. Do you want to exclude the axiom of regularity too? – Johannes Hahn Aug 22 '18 at 17:25
• Yes of course, no regularity is assumed. Let me correct that – Zuhair Al-Johar Aug 22 '18 at 17:28
• I think that you can still form the set's Hartogs number, which would be in the cumulative hierarchy. – James Hanson Aug 22 '18 at 17:52
• @ZuhairAl-Johar: As James points out, given a set $X$, you can show the existence of an ordinal which does not inject into $X$ in ZF-Regularity. Perhaps you might want to exclude replacement as well? – Burak Aug 22 '18 at 18:11
• @JamesHanson, Yes, you are right, then the base theory must be a fragment of ZF that doesn't prove existence of a Hartog number for any set. Ok, I'll edit it. – Zuhair Al-Johar Aug 22 '18 at 18:55

It is consistent with ZF-Regularity to have a class $Q$ of Quine atoms of any size, even we can have a proper class of Quine atoms, so let $$|Q|=|V_{\omega+\omega}|$$, we simply construct the set $V^Q_{\omega+\omega}$, that is the iterative hierarchy having $\omega+\omega$ stages with $V_0 = Q$, and this would give a positive answer to this question.
I think that this simple argument not only work for $Z$ but for any fragment of $ZF$ with some cardinal limit on Replacement, so this I think would stop the hartog ordinal from being constructed from the the big set of this question. That said, then this argument can even work with choice.