0
$\begingroup$

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 $V_{\omega+{\omega}}$

$\endgroup$
  • 2
    $\begingroup$ Wait, even without Choice, every set is itself somewhere in the cumulative hierarchy. Do you want to exclude the axiom of regularity too? $\endgroup$ – Johannes Hahn Aug 22 '18 at 17:25
  • $\begingroup$ Yes of course, no regularity is assumed. Let me correct that $\endgroup$ – Zuhair Al-Johar Aug 22 '18 at 17:28
  • 3
    $\begingroup$ I think that you can still form the set's Hartogs number, which would be in the cumulative hierarchy. $\endgroup$ – James Hanson Aug 22 '18 at 17:52
  • 2
    $\begingroup$ @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? $\endgroup$ – Burak Aug 22 '18 at 18:11
  • $\begingroup$ @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. $\endgroup$ – Zuhair Al-Johar Aug 22 '18 at 18:55
0
$\begingroup$

I think I have the answer to this question, the general idea I got from Noah Schweber's prior answer.

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.