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Consider the definition of measurable cardinal (this definition was found in Neil Barton's paper, "Large cardinals and the iterative conception of set"):

Definition 8. A cardinal $\kappa$ is measurable iff it is the critical point of a non-trivial elementary embedding $j$: $V$ $\rightarrow$ $M$, for some transitive inner model $\mathfrak M$ = ($M$, $\in$ ).

Given my limited understanding of set-theoretic potentialism, the above definition seems to me to rest on actualist assumptions (in particular, the assumption that $V$ actually exists).

This interpretation is born out of my understanding of current writings of set-theoretic potentialists, in particular, the slide presentation of Joel David Hamkins titled "Potentialism and implicit actualism in the foundations of mathematics" (his Jowett Society lecture, Oxford, February 2019). Consider the following, from his slide presentation:

A potentialist system $\mathcal W$ is convergent, with limit $M$, if

Every world in $\mathcal W$ is a substructure of $M$.

Every world in $\mathcal W$ can be extended so as to accomodate any desired individual of $M$.

The archetypal limit structure would (at least to actualists) at first glance seem to be $V$. However, Prof. Hamkins in another slide, states the following:

Potentialist translation

For every $\psi$ in $\mathcal L$, form the potentialist translation $\psi^{\lozenge}$ by

replace $\exists$$x$ by $\lozenge$$\exists$$x$ replace $\forall$$x$ by $\Box$$\forall$$x$.

Theorem. If a potentialist system $\mathcal W$ has has limit $M$ then

$M$ $\vDash$ $\psi$($a$) $\Longleftrightarrow$ $W$ $\vDash_{\mathcal W}$ $\psi^{\lozenge}$($a$),

for any world $W$ $\in$ $\mathcal W$ in which the individual $a$ exists.

Thus, actual truth in the limit structure amounts to potentialist truthe in the approximating structures. So the potentialist can in fact refer to th actual truth.

Proved by simple induction on formulas.

Finally, he writes the following:

The potentialist translation shows that in convergent potentialism, the potentialist can nevertheless accurately refer to truth in the limit structure, which is not part of the potentialist ontology.

Convergent potentialism is thus a form of implicit actualism.

Although the actual limit structure does not exist, nevertheless one has a full account of the objects and truths of it in the potentialist language. It is as though the actual limit structure exists in all the ways that matter.

But is this sufficient enough to derive the notion of measurable cardinal as stated above?

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    $\begingroup$ I honestly don't understand what this is asking. The text you've cited describes a way to translate actualist accounts into potentialist accounts; what does that have to do with a measurable cardinal specifically? Are you asking whether a particular actualist argument for measurables can be translated into a potentialist argument for measurables? If so, what argument do you have in mind? $\endgroup$ Commented Feb 27, 2019 at 23:20
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    $\begingroup$ I agree that the question is not particularly clear, but one way of making sense of it would be to ask whether one necessarily needs to talk about all of $V$ in order to define measurability, or do bounded pieces of the universe suffice (I suppose "bounded" can be interpreted in different ways, depending on the kind of potentialism one is interested in). If you allow pieces of the from $V_\alpha$, then the answer is yes, as the measurability of $\kappa$ is witnessed by objects somewhere in the neighbourhood of $V_{\kappa+3}$ (either the measure, or just an embedding defined on $V_{\kappa+1}$). $\endgroup$ Commented Feb 28, 2019 at 8:57
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    $\begingroup$ If you restrict your pieces of the universe to have size at most $\kappa$, then I suspect you won't be able to express measurability, but will get stuck somewhere in the region of a Ramsey cardinal. $\endgroup$ Commented Feb 28, 2019 at 8:59
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    $\begingroup$ But I don’t understand what you mean by “derive the notion of measurable cardinal.” $\endgroup$ Commented Feb 28, 2019 at 15:55
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    $\begingroup$ @ThomasBenjamin Why doesn't the (original!) definition "$\kappa$ is measurable iff there is a $\kappa$-complete nonprincipal ultrafilter on $\kappa$" count as "local enough" for your purposes? It's completely first-order, and pays attention only to a very bounded amount of the universe (namely, whether $\kappa$ is measurable or not is determined entirely by $V_{\kappa+2}$). Indeed, these sort of "small generators" for elementary embeddings are crucial tools for understanding these more complicated (and "class-flavored") large cardinal notions ... $\endgroup$ Commented Mar 7, 2019 at 16:49

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Paring the philosophy away (which I think just obscures things) this seems to just be a question of definability - the key point being that Hamkins' recursively-defined translation works for any formula in the language of set theory. The elementary embeddings definition of measurability isn't such a formula since it involves quantification over classes, and so you may ask:

Is there a way to define measurability in ZFC alone? (And if so, how do we argue that this definition and the elementary embeddings definition actually coincide?)

  • Note that this is an important question independent of your stance on set-theoretic (non-)reality. This is why I think it's valuable to pare away the philosophy as much as possible: there's a perfectly fine mathematical question here, and throwing philosophy into the mix just makes it harder to discern.

Such a definition does in fact exist, and was how measurability was defined originally. Namely, a cardinal $\kappa$ is measurable iff there is a nonprincipal $\kappa$-complete ultrafilter on $\kappa$. This is a purely combinatorial definition, and whether $\kappa$ is measurable is determined entirely by $V_{\kappa+2}$ - we don't need to "look at all of $V$."

Now, in what sense can we say that this definition equivalent to the elementary embeddings one? Well, what we do is pass to NBG, which can make perfect sense of both definitions and conservatively extends ZFC, and prove in NBG that the two definitions are equivalent.


Other large cardinals have similar translations. For example, supercompactness has a "class-wise" definition

$\kappa$ is supercompact iff for every ordinal $\lambda$ there is some inner model $M$ with $M^\lambda\subseteq M$ and some elementary embedding $j_\lambda: V\rightarrow M$ with $crit(j)=\kappa$ and $j(\kappa)>\lambda$.

and a "set-wise" one

$\kappa$ is supercompact iff for every $\mu\ge\kappa$ there is a normal measure over $[\mu]^{<\kappa}$.

Again, the latter definition just makes reference to sets, and the equivalence of the two definitions is proved in NBG (where they each make sense).

At this point it's a good idea to try to translate supercompactness into the potentialist language without using Hamkins' explicit translation, to get a feel for how things relate. We reason as follows:

  • First, suppose $\mathcal{W}$ had a limit $V$ and $\kappa$ were supercompact in $V$. Then for every $\mu\ge\kappa\in V$ there must be a $u_\mu$ in $V$ which is a normal measure on $[\mu]^{<\kappa$}$ ... all in the sense of $V$.

  • OK, now since $V$ is the limit of $\mathcal{W}$, everything in $V$ shows up in some world in $\mathcal{W}$. Obviously this includes $\kappa$ itself - the definition of "$\kappa$ is a $\mathcal{W}$-supercompact" needs to begin with "for some $M\in\mathcal{W}$, $\kappa\in M$" - but also the $\lambda$s and $u_\lambda$s will "show up along the way."

  • ... But maybe not at the same time. Whenever we have quantifier alternation we have to "move along $\mathcal{W}$" - e.g. the definition of "For all $\lambda\ge\kappa$ there is a $u_\lambda$ such that ..." needs to begin "For each world $N\supseteq M$ and each $\lambda\ge\kappa$ in $N$, there is some further world $K\supseteq N$ and some $u_\lambda\in K$ such that ..."

  • Continuing in this way you wind up with a definition of "$\mathcal{W}$-supercompact" which works - that is, such that in any potentialist system $\mathcal{W}$ which has a limit $W$, "$\mathcal{W}$-supercompact" and "supercompact$^W$" are equivalent. But more than this, at this point the fully general potentialist translation Hamkins gives should be clear.

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Yes, it is sufficient. Your question actually touches on a known, similar issue. A proper class is not an object in ZF, so in ZF we can't directly refer to it. Hence, again under the assumption we work in ZF, $V$ is not an object, $j$ is not an object, and if $M$ is a proper class, $M$ is also not an object. Despite all this, mathematicians working in ZF talk about these things indirectly.

The trick is, one can assume, with an exception to the case where $j \colon V \to V$, that classes $M$ and $j$ are definable from parameters by $\Sigma_2$ formulas. Moreover, the assertion that $j$ is an elementary embedding is equivalent to the assertion that for all formulas $\varphi(x)$, all ordinals $\alpha$, and all sets $a \in V_\alpha$ $$V_\alpha \models \varphi[a]$$ if and only if $j(V_\alpha) \models \varphi[j(a)]$.

So, getting back to your original question, the answer is yes. One can talk about an elementary embedding $j \colon V \to M$ without accepting the existence of $V$. The method is the same as the one outlined above.

Another way of looking at your problem is to think of a less complicated, similar scenario. In classical potentialism, one refutes the existence of $\mathbb{N}$ but accepts all its finite initial segments. Here, the problem would be whether one can talk about functions $f \colon \mathbb{N} \to A$, where $A$ is some finite set of natural numbers. In particular, let us call $n$ a pigeonhole number whenever there is an $m<n$ such that $f(m) = f(n)$.

The question: For every finite set of natural numbers $A$, does every function $f \colon \mathbb{N} \to A$ admit a pigeonhole number? While $f$ and $\mathbb{N}$ don't exist for a classical potentialist, he can still answer this question! He can't do it explicitly, but he can do it implicitly using larger and larger finite initial segments of $\mathbb{N}$.

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    $\begingroup$ What is an "object"? Why is the set of all real numbers an "object" while the class of all ordinals is not an "object"? $\endgroup$
    – Nik Weaver
    Commented Mar 5, 2019 at 22:14
  • $\begingroup$ @Wojciech: Welcome to MathOverflow! How (by the way) does your answer relate to Miha's comment (if at all)? $\endgroup$ Commented Mar 6, 2019 at 20:49
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    $\begingroup$ @NikWeaver if you take first order ZF seriously as your foundational theory, then this follows. $\endgroup$ Commented Mar 7, 2019 at 6:09
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    $\begingroup$ @MonroeEskew: if you take NBG as your foundational theory it is wrong. This is not a serious philosophical argument. $\endgroup$
    – Nik Weaver
    Commented Mar 7, 2019 at 11:37
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    $\begingroup$ What concerned me about the original answer was that it seemed to be making a global statement ("$V$ is not an object ... regardless of our philosophical inclinations"). $\endgroup$
    – Nik Weaver
    Commented Mar 7, 2019 at 16:45

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