# Is the power set axiom essential for constructing L?

Take ZFC, remove axiom of Power set, and put instead of it the following axiom:

Axiom of Successor Cardinals: $$\forall \kappa \exists x \forall \alpha ( \alpha \leq \kappa \to \alpha \in x)$$

where "$$\leq$$" refers to "cardinal smaller than or equal" relation, and $$\kappa, \alpha$$ range over von Neumann ordinals.

Can the resulting theory still interpret ZFC?

The idea is that if we can develop Gödel's constructible universe L inside this system, then this would interpret ZFC? So the power set axiom won't be essential for the development of L?

• But $\aleph_{\omega}$ is already a set in this theory so how can it be interpreted in $L_{\aleph_{\omega}}$ – Zuhair Al-Johar Aug 13 '19 at 14:14
• The title is misleading, since it implies you're thinking about ZFC-, rather than ZFC- with additional axioms. Your question, phrased differently, I think, is if we close the ordinals under Cartesian products, unions, and Replacement, do we get L? – Asaf Karagila Aug 13 '19 at 14:25
• Since KP alone proves that $L_\alpha$ exists for every ordinal $\alpha$ (appropriately phrased), we can indeed "build $L$" inside this theory. Essentially the usual argument then shows that the resulting structure satisfies full ZFC (+ V=L). – Noah Schweber Aug 13 '19 at 14:57
• @AsafKaragila I agree that the title is somehow misleading, but the idea is that cartesian products and successor cardinals are in some sense weaker than power set axiom, since they are implied by the power set axiom, but the converse is not true, i.e. the power set axiom is not a theorem of the above mentioned theory. I think if one read my message then it is clear that I'm asking about ZFC minus power set plus axioms of Cartesian products and Successor cardinals (you seem to forget about the last axiom). – Zuhair Al-Johar Aug 13 '19 at 15:27
• @ZuhairAl-Johar Actually KP alone does it. – Noah Schweber Aug 13 '19 at 16:56

KP alone - which is vastly weaker than the theory in question - proves the sentence "For every ordinal $$\alpha$$, $$L_\alpha$$ exists," since it is strong enough to enable effective transfinite recursion. (We're passing to an unnecessarily weak subtheory, but it's worth noting.) The proof of this can be found e.g. in Barwise's book.

The condensation lemma, appropriately stated, can also be proved in KP; since our theory proves that successor cardinals exist, we get powerset in $$L$$. The proof that $$L$$ satisfies the rest of the ZFC axioms is the usual one.

So there is a uniform way to define in an arbitrary model $$M$$ of your theory an inner model (= transitive subclass containing all the ordinals of $$M$$) which is a model of ZFC + V=L.

Note the role of condensation in the above: condensation reduces powersets in $$L$$ to successor cardinals in $$L$$ (and hence a fortiori in any larger class). So it's not so much that we're avoiding powerset in building $$L$$, but rather that a very weak theory proves that powerset-in-$$L$$ is equivalent to successor-cardinals-in-$$L$$.

• So this proves that we don't need to axiomatize power-set in order to build $L$. Axiom of Successor cardinals is weaker than axiom of power set, and yet it manages to build $L$ and so its consistency proves the consistency of ZFC, i.e. of adding power to the rest of axioms of ZFC. In nutshell ZFC-power+Successor cardinals can work as a foundational theory much as ZFC works, because simply its consistency proves the consistency of ZFC. In some sense Power set axiom is bypassed. – Zuhair Al-Johar Aug 14 '19 at 10:28
• @ZuhairAl-Johar Your philosophical claim has a big underlying assumption - that the "foundational satisfactoriness" of a theory is entirely determined by in its interpretability strength. Why shouldn't it hinge on what the theory actually does or does not prove outright? Also, it's not clear to me that this is a persistent phenomenon - what do we add to this theory to get something which interprets (say) ZFC + "There is a supercompact cardinal"? I think the above idea doesn't work here to show that we can more-or-less add the same sentence verbatim since we don't have a good fine structure ... – Noah Schweber Aug 14 '19 at 10:32
• Actually you seem to be doing something even more restrictive, and focusing on consistency strength. Would you consider I$\Sigma_1$ + "ZFC is consistent" - which is even stronger than ZFC in consistency strength - to be foundationally satisfying? What about I$\Sigma_1$ + "ZFC is consistent" + "I$\Sigma_1$ + "ZFC is consistent" is inconsistent"? – Noah Schweber Aug 14 '19 at 10:33
• @of course the consistency strength cannot be the sole criterion for a foundational status of a theory. The theories you've mentioned already speak of ZFC, so they are not bypassing ZFC conceptually speaking. I agree with the extendability point you've mentioned. But those can be done by extracting ZFC as a byproduct of this theory, and then working on extending it. The point is that we don't need to axiomatize power set to get to ZFC. However axiom of successor cardinals seem to be necessary to get to ZFC via V=L. Actually I'm aiming at a more reductive measure, that of ,.to be continued – Zuhair Al-Johar Aug 14 '19 at 12:39
• ..continuation: .. eliminating the hierarchical set structure altogether and collapsing it to a flat set structure of ordinals, as to get a simple theory of ordinals and sets of them that encodes relations between them as I've posted to Mathoverflow at a prior posting, which I think we can build L within it and interpret ZFC. That would be a true reduction in structure that deserves a foundational status. – Zuhair Al-Johar Aug 14 '19 at 12:43