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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.

Note: the relation $\leq$ is defined as:

$\alpha \leq \beta \iff \exists f \, (f:\alpha \to \beta \land f \text { is an injection})$

so, it need not be confused with ordinal smaller than or equal relation.

So, the axiom is saying that for every cardinal $\kappa$, there is a successor cardinal $\kappa^+$.

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?

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  • $\begingroup$ But $\aleph_{\omega}$ is already a set in this theory so how can it be interpreted in $L_{\aleph_{\omega}}$ $\endgroup$
    – Zuhair Al-Johar
    Commented Aug 13, 2019 at 14:14
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    $\begingroup$ 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? $\endgroup$
    – Asaf Karagila ♦
    Commented Aug 13, 2019 at 14:25
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    $\begingroup$ 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). $\endgroup$
    – Noah Schweber
    Commented Aug 13, 2019 at 14:57
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    $\begingroup$ @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). $\endgroup$
    – Zuhair Al-Johar
    Commented Aug 13, 2019 at 15:27
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    $\begingroup$ @ZuhairAl-Johar Actually KP alone does it. $\endgroup$
    – Noah Schweber
    Commented Aug 13, 2019 at 16:56

3 Answers 3

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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-$L$.

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  • $\begingroup$ 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. $\endgroup$
    – Zuhair Al-Johar
    Commented Aug 14, 2019 at 10:28
  • $\begingroup$ @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 ... $\endgroup$
    – Noah Schweber
    Commented Aug 14, 2019 at 10:32
  • $\begingroup$ 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"? $\endgroup$
    – Noah Schweber
    Commented Aug 14, 2019 at 10:33
  • $\begingroup$ @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 $\endgroup$
    – Zuhair Al-Johar
    Commented Aug 14, 2019 at 12:39
  • $\begingroup$ ..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. $\endgroup$
    – Zuhair Al-Johar
    Commented Aug 14, 2019 at 12:43
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Let me not answer the question asked but add an important angle.

An $L$ can be built already in $\mathsf{ATR}_0$, which is the weakest theory that can do it convincingly (some coding involved).

You can't guarantee Powerset in that $L$, but it can well happen that this $L$ acquires lots of uncountable cardinals. (All ordinals were "countable" in the initial model of arithmetic, but after the extraction of $L$, many original bijections between ordinals and $\mathbb N$ were left outside.)

I guess the best source for this is Simpson's "Subsystems of Second Order Arithmetic", parts VII.3 and VII.4.

Perhaps what is also very relevant to your thoughts is something called "the Feferman-Leví model" discussed on page 295 of Simpson's book and elsewhere.

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Let me address the question in the body of the post, rather than the title question. Namely, you asked whether your theory interprets ZFC.

Negative answer with axiom as stated. The natural reading of your axiom $\forall\kappa\exists x\forall\alpha(\alpha\leq\kappa\to x\in\alpha)$ in set theory would be that it asserts the existence of a set $x$ containing every ordinal $\alpha$ with $\alpha\leq\kappa$, which is to say, it asserts the existence of the ordinal successor $\kappa+1$.

On this reading of your axiom, the answer is negative, the theory would not interpret ZFC. As the other answers have noted, it can build $L$, but the $L$ that it builds will not necessarily be a model of ZFC.

The reason is that the theory ZFC-P already proves the existence of ordinal successors, but it is too weak to interpret ZFC. This is a consequence of the fact that ZFC proves Con(ZFC-P), since ZFC proves that the structure of hereditarily countable sets HC is a model of ZFC-P. But ZFC-P cannot interpret a theory that proves Con(ZFC-P) by the second incompletness theorem.

More concretely, if one starts in the ZFC-P model HC of hereditarily countable sets, then the $L$ that one builds will be exactly $L_{\omega_1}$, which is a model of $V=L$, but it is not necessarily a model of ZFC.

Positive answer with corrected axiom. You have clarified in the comments, however, and in the post that you have an idiosyncratic meaning for $\leq$ in your axiom and that you intend it to assert the existence of cardinal successors. That is, we are working in ZFC-P + every cardinal $\kappa$ has a successor cardinal $\kappa^+$.

In this case, the answer becomes affirmative. As we noticed, ZFC-P can construct the inner model $L$, and if every cardinal has a successor, then those cardinals will arise inside $L$ and so $L$ will also think that every cardinal has a successor. And it will think that the replacement axiom is true, if replacement holds in $V$. We will get the power set axiom being true inside $L$ because the subsets of a set show up in the $L$ hierarchy before the next cardinal stage, and so we can collect them into a set by applying comprehension to that stage $L_\gamma$. In this way, all the ZFC axioms will be true in the $L$ that we build, and so we are interpreting ZFC.

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  • $\begingroup$ But $\leq$ refers to cardinal smaller than or equal relation and not to ordinal smaller than or equal relation. So, I think the axiom as I've stated it does capture successor cardinals and cannot be captured by successor ordinals. $\endgroup$
    – Zuhair Al-Johar
    Commented Aug 27, 2023 at 13:39
  • $\begingroup$ That is very idiosyncratic notation. I had thought you had meant to restrict to cardinals $\alpha$, in which case the order relation is the same as for ordinals. But if you meant $|\alpha|\leq\kappa$, then you are indeed asserting that $\kappa^+$ exists. But why not say it that way? Every cardinal has a successor cardinal. It would be much clearer. $\endgroup$
    – Joel David Hamkins
    Commented Aug 27, 2023 at 14:00
  • $\begingroup$ Your $\leq$ notation is very bad, since you will end up asserting things like $\omega+5\leq\omega$ and so forth. For ordinals $\alpha,\beta$, the notation $\alpha\leq\beta$ is universally understood to mean the ordinal order, and you invite needless confusion by redefining this totally standard relation. It is needless because you could just write $|\alpha|\leq\beta$ for your intended meaning. But it is better to just formulate your intended axiom by saying: every cardinal $\kappa$ has a successor cardinal $\kappa^+$. $\endgroup$
    – Joel David Hamkins
    Commented Aug 27, 2023 at 14:09
  • $\begingroup$ Ok, then I'll re-write it, I thought if I just say "cardinal" smaller than or bigger relation, then I'm done, and I thought it was clear. Thanks. $\endgroup$
    – Zuhair Al-Johar
    Commented Aug 27, 2023 at 14:12
  • $\begingroup$ The cardinal order relation on the set of cardinals is the same as the ordinal relation on the set of cardinals. But you are using the cardinality order relation on the set of ordinals, which is weird. No need to edit, though, since the question already has answers, and it will perhaps confuse the discussion to change it. $\endgroup$
    – Joel David Hamkins
    Commented Aug 27, 2023 at 14:16

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