1
$\begingroup$

In ZFC, the following holds: If $\kappa$ is an infinite cardinal, then $\alpha \cdot \beta < \kappa$ for all $\alpha, \beta < \kappa$.
How much of ZFC can we remove without losing this property? It seems that the Axiom of power set is not necessary. Is replacement necessary?

$\endgroup$
  • 1
    $\begingroup$ You can remove choice : ZF alone is sufficient for this $\endgroup$ – Max Sep 15 '17 at 22:39
  • $\begingroup$ After thinking a bit about it, I agree with you that choice is not necessary. Trivially, we can remove the axiom of infinity. The axiom of power set seems to be superfluous, because we don't need it in the definition of ordinal multiplication or in the proof of the above property. I'm not sure about replacement. $\endgroup$ – User_754 Sep 16 '17 at 23:38
  • 1
    $\begingroup$ You do not need replacement either. $\endgroup$ – Andrés E. Caicedo Sep 17 '17 at 2:25
  • $\begingroup$ So this works in ZF without the axiom of infinity, replacement and the axiom of power sets. What if we remove separation from this theory? Intuitively, I would say that it doesn't work then. $\endgroup$ – User_754 Sep 17 '17 at 9:26

Your Answer

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

Browse other questions tagged or ask your own question.