Ordinal numbers are generalizations of natural numbers. In this sense the "proper class" of all ordinals ($Ord$) is very similar to "infinite" set of all natural numbers ($\omega$). In the other direction we know that many large cardinal axioms are generalizations of the properties of $\omega$ and without assumption of uncountablity one can prove (in $ZFC$) that $\omega$ is a large cardinal in many types. For example $\omega$ is a strongly compact cardinal because $\mathcal{L}_{\omega , \omega}$ is a compact logic. This fact shows that the nature of $\omega$ is very adequate to be a large cardinal. Now a natural question arises for the $Ord$ like this:

Question (1): Is the nature of $Ord$ adequate to be a large cardinal? More precisely if $A$ be a large cardinal type, then how strong is the statement: "$Ord$ is a large cardinal of type $A$"?

It seems that because of the similarity between $\omega$ and $Ord$ these statements must be very close to a provable statement in $ZFC$ or be a "very weak large cardinal axiom". For example it is provable in $ZFC$ that "$Ord$ is strongly inaccessible" and the statement "$Ord$ is Mahlo" is weaker than existence of an "uncountable" Mahlo cardinal (see the Cantor's upper attic for more details). Even there are some problems during investigating large cardinal properties of $Ord$, for example in defining some types of large cardinality for $Ord$ particularly in large cardinals based on elementary embedding definitions. So:

Question (2): Is there an equivalent definition for any large cardinal axiom $A$ which the statement "$Ord$ is a large cardinal of type $A$" be meaningful?

Question (3): Is there a large cardinal axiom like $A$ such that: $ZFC\vdash \neg~(Ord~is~a~large~cardinal~of~type~A)$

  • $\begingroup$ For posterity, a more recent result (which OP has commented on) is that ZFC proves "Ord is not definably weakly compact", with the restriction to definable proper classes being necessary in first-order ZFC (much the same way "Ord is Mahlo" is formulated). jdh.hamkins.org/ord-is-not-definably-weakly-compact $\endgroup$
    – C7X
    Commented Jan 18, 2023 at 3:19

1 Answer 1


$\newcommand{\Ord}{\text{Ord}} \newcommand{\ZFC}{\text{ZFC}}$

Here is one way to formalize your concept a little more tightly, which provides the answers to your questions. For any large cardinal property $P$, let's take the phrase "$\Ord$ is $P$" to be the theory asserting $\sigma$, for any sentence $\sigma$ that ZFC proves is true in $V_\delta$ under the assumption that $\delta$ has property $P$ in $V$.

With this formalization, "$\Ord$ is $P$" asserts that the universe is just like we would expect, if we were living inside $V_\delta$ for an actual $P$ cardinal $\delta$. For example, "$\Ord$ is measurable" implies that there are a proper class of weakly compact cardinals, since this is true in $V_\delta$, whenever $\delta$ is measurable, and "$\Ord$ is supercompact" implies that there are many partially supercompact cardinals, with nice limit properties. For example, they would form a stationary class in the sense that every definable class club would contain one of them. This notion seems to capture what one would want to mean by saying $\Ord$ has property $P$ as a purely first-order theory about sets.

With this idea, the point I would like to make is that assuming $\Ord$ is $P$ is essentially equivalent to assuming that what you have is $V_\delta$, where $\delta$ has property $P$ in a larger universe.

Theorem. For any large cardinal property $P$, a model of set theory $M$ satisfies "$\Ord$ is $P$" if and only if $M\prec V_\delta^N$ for some taller model of set theory $N$ with a cardinal $\delta$ having property $P$ in $N$.

Proof. The backward direction is immediate, since $\delta$ having property $P$ implies that $V_\delta$ satisfies every assertion of $\Ord$ is $P$. For the forward direction, suppose $M$ satisfies $\Ord$ is $P$. Let $T$ be the theory consisting of $\ZFC$, plus the assertion "$\delta$ is $P$", using a new constant symbol $\delta$, plus the assertions $\varphi^{V_\delta}$, for any $\varphi$ in the elementary diagram of $M$, using constants for elements of $M$. This theory is finitely consistent, since otherwise there would be finitely many assertions in it that are contradictory, and so there would be a statement $\varphi$ true in $M$ that provably could not hold in $V_\delta$ for any cardinal $\delta$ with property $P$. But that would contradict our assumption that $M\models\Ord$ is $P$.

If $N$ is any model of the theory, then $\delta$ has property $P$ in $N$, and we get $M\prec V_\delta^N$, because $V_\delta^N$ satisfies the elementary diagram of $M$. Another way to say this is that there is an elementary embedding $j:M\to V_\delta^N$, mapping every element of $M$ to the interpretation of its constant in $N$. QED

Thus, if one is inclined to assume $\Ord$ is $P$, then why not go ahead and make the full move to a model with an actual $P$ cardinal $\delta$, such that our old world looks exactly like $V_\delta$ in this new world. In particular, under this terminology, the theory $\ZFC+\Ord$ is $P$ is equiconsistent with $\ZFC+\exists \delta$ with property $P$.

Corollary. The following theories are equiconsistent:

  1. $\ZFC+\Ord$ is $P$.
  2. $\ZFC+\exists \kappa$ with property $P$.

Lastly, I would like to point out that there is some variance in the literature about what "$\Ord$ is $P$" should mean. For example, one often finds the phrase "$\Ord$ is Mahlo" to mean only the weaker assertion, that every definable closed unbounded class of cardinals contains a regular cardinal. This is what one finds, for example, at Cantor's Attic. But this is strictly weaker in consistency strength than ZFC+$\exists\kappa$ Mahlo, since this latter theory implies the consistency of the former, as it is true in $V_\kappa$ whenever $\kappa$ is Mahlo.

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    $\begingroup$ This answer, however, does not address the part of your question about properties of Ord generalizing properties of $\omega$, which is an interesting topic about which people have said quite a lot. Perhaps someone will post an answer along those lines. $\endgroup$ Commented Sep 3, 2013 at 15:37
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    $\begingroup$ Something doesn't sit right with this corollary. In every model of $\sf ZFC$ we have that $\sf Ord$ is strongly inaccessible. $\endgroup$
    – Asaf Karagila
    Commented Sep 3, 2013 at 16:24
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    $\begingroup$ I don't agree that it is right to say that ZFC proves that Ord is inaccessible, but rather only that it is "definably inaccessible" or "definably regular", in the sense that it proves only that no definable class is a singularizing function. This is strictly weaker than actual regularity, as witnessed by the singular worldly cardinals. ZFC proves really only that "Ord is worldly". $\endgroup$ Commented Sep 3, 2013 at 18:15
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    $\begingroup$ My view is that the slogan "ZFC proves that Ord is inaccessible" is just a metaphor, given to understand what we are trying to get at with the replacement axiom. A more careful view leads to the definition I give here, and with this view we shouldn't really say that ZFC proves Ord is inaccessible. $\endgroup$ Commented Sep 3, 2013 at 19:02
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    $\begingroup$ I think it is worth mentioning that the "taller model $N$" mentioned in your theorem is in general not well-founded. (Not that there is anything wrong with that!) $\endgroup$
    – Goldstern
    Commented Sep 3, 2013 at 21:37

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