Does the fact that, assuming the consistency of $ZFC$, no proof that the consistency of "$ZFC$ implies the consistency of '$ZFC$ + There exists a weakly inaccessible cardinal'" can be formulated in $ZFC$ (this paraphrased from the Wikipedia entry, "Inaccessible cardinal") imply that, given a model $M$ of $ZFC$ in which no weakly inaccessible cardinal exists, there can be no generic extension $M[G]$ of $M$ in which a weakly inaccessible cardinal was forced to exist?

I conjecture that the answer to this question is 'yes' due to the answers given to Ewen Delanoy to his mathstackexchange question, Relation between inaccessible cardinals and $CH$. The answers were given by arjafi and Asaf Karagila. I quote the relevant paragraph of Asaf's answer first (my comments will be in square brackets throughout):

Do note, however, that the assumption of $CH$ is equiconsistent with its negation, as Godel's and Cohen's work show; while this is not true for the existence of large cardinals. The assertion that there exists a [weak] inaccessible is strictly stronger than the assertion that there are none [this asymmetry (I believe) implies that there is no notion of forcing that will produce a generic extension $M[G]$ of a model $M$ of $ZFC$ containing no weakly inaccessible cardinal such that $M[G]$ contains a weak inaccessible since forcing is essentially a relative consistency proof, i.e., that given a model $M$ of $ZFC$ ($CON(ZFC)$), there is a forcing extension $M[G]$ in which $\varphi$ holds ($CON($ZFC + ${\varphi}$))].

The reason for this asymmetry is described by arjafi in his answer:

This is where we have to hedge a bit. It is actually unknown whether any large cardinal assumptions are relatively consistent with $ZFC$. Furthermore, we cannot prove that large cardinals are relatively consistent with $ZFC$ without transcending $ZFC$ itself; that is proving the relative consistency from a metatheory stronger than $ZFC$.

So the following question remains (assuming the answer to my first question is 'yes'):

"What is this metatheory?" (Another way of possibly rephrasing this question is: "What strong base theory $T$ in the language of set theory (could be second-order or greater) which does not assume outright the existence of an inaccessible cardinal can have (say) $PRA$ proves "$CON$($T$ + There exists a weakly inaccessible cardinal) $\Leftrightarrow$ $CON$( $T$ + There is no weakly inaccessible cardinal)".) This state of affairs (I think) would allow one to have notions of forcing which, from a model $M$ of "$T$ + 'There is no weakly inaccessible cardinal'" one could construct a generic extension $M[G]$ satisfying "$T$ + 'There exists a weakly inaccessible cardinal'".

A natural candidate for $T$ might be $KM$ (Kelly-Morse) set theory, but as correctly argued in Joel David Hamkins' blogpost "Kelly-Morse set theory implies $CON(ZFC)$ and much more":

...the consistency strength of $KM$ lies strictly that of $ZFC$, above much of the iterated consistency hierarchy, but below that of $ZFC$ plus an inaccessible cardinal.

so $T$ $\ne$ $KM$.

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    $\begingroup$ The answer to your first question is yes, because inaccessibility is downwards absolute. In other words, if $\kappa$ is inaccessible in $V$, then it is inaccessible in any inner model, in particular in any putative ground model of $V$. But note that the situation is less straightforward for other large cardinals. For example, it is possible to force to make a nonmeasurable $\kappa$ become measurable in an extension (indeed, this can happen after forcing with $\operatorname{Add}(\kappa,1)$). $\endgroup$ Oct 26, 2017 at 7:53
  • $\begingroup$ @MihaHabič: Thanks, but in the case where one is forcing a nonmeasurable $\kappa$ to be measurable in the extension $M[G]$, $\kappa$ must be inaccessible in $M$, correct (by downward asoluteness) ? $\endgroup$ Oct 26, 2017 at 11:07

1 Answer 1


Before leaping into my rather long answer, the three-sentence response to your question is: since strengthening the theory only limits the class of models, you'll never make something forcible that was never forcible before by passing to a stronger theory. You can never force an inaccessible over a model of ZFC, so replacing ZFC with a stronger theory won't change the situation. In particular, mathematical objects (like large cardinals) are fundamentally more than just the consistency strength (or other logical strengths) they carry.

OK, now on to the full answer ...

First of all, it is true that no forcing can make a non-inaccessible cardinal inaccessible, and as Miha says in his comment in fact more is true: if $M, N$ are models of ZFC with $M\subseteq_{end} N$ (that is, $N$ is an end extension of $M$: if $x\in M$, $y\in N$, and $y\in^Nx$, then $y\in M$; don't confuse this with the notion of a "top extension"), then every inaccessible in $N$ is also inaccessible in $M$. Similarly, every weakly compact cardinal in $N$ is weakly compact in $M$, etc.

(To address your question in your response to Miha's comment, yes: since measurable implies inaccessible, anything which is "forcibly measurable" must already by inaccessible. Similarly, since measurable implies weakly compact, anything which is "forcibly measurable" must already be weakly compact. This line of reasoning applies to inaccessiblity and weak compactness since they are downwards absolute; by contrast, similar arguments would fail for more complicated large cardinals - e.g. this argument does not immediately show that a "forcibly Woodin" cardinal must already be measurable.)

That said, the rationale you give for this in the first paragraph is not correct. For example, the following is true:

$(*)\quad$ Assuming the consistency of ZFC + Con(ZFC) (this last piece was missing in your statement, and when I first wrote this answer I missed it myself), no proof that the consistency of "ZFC implies the consistency of 'ZFC + There exists a weakly inaccessible cardinal'" can be formulated in ZFC.

(Although I think it flows better to say "Assuming ZFC + Con(ZFC) is consistent, ZFC cannot prove "If ZFC is consistent then ZFC+an inaccessible is consistent.")

From now on I'll sweep the usual meta-commitments like "ZFC + Con(ZFC) is consistent" under the rug.

Yet we can have "forcibly measurable" cardinals which are not measurable. So this shows that it's not enough to simply observe that a large cardinal assumption has high consistency strength in order to be certain it can't be forced.

Of course, what we can conclude from this situation is that such a principle can't always be forced:

$(**)\quad$ Suppose ZFC+$\varphi$ proves the consistency of ZFC. Then there is a countable model $M$ of ZFC such that no forcing extension of $M$ satisfies $\varphi$ (assuming ZFC is consistent of course).

(Note that forcing extensions of countable structures exist outright, so there's no metamathematical issues related to forcing in the above statement.)

To prove $(**)$, note that otherwise by the completeness theorem we would have that ZFC proves "There is some $\mathbb{P}$ such that $\mathbb{1}_\mathbb{P}\Vdash_\mathbb{P}\varphi$." This gives in turn a proof that from a model of ZFC we can produce a model of ZFC+$\varphi$, and this proof goes through in an appropriately weak theory. (OK, I phrased this claim in terms of models, so this "appropriately weak" theory still needs to be able to talk about sets and structures; we can bring this down to PRA as usual, but that's a bit tedious.)

Now when you ask about the base theory, this same confusion between consistency and forcing crops up in your expectation that higher consistency strength implies the potentiality of large cardinals. Restricting attention to an extension $T$ of ZFC for the moment for simplicity, the point is that the same proof that inaccessibility is downwards absolute between end extensions of ZFC shows a fortiori that inaccessibility is downwards absolute between end extensions of models of $T$. Adding strength makes it harder to force large cardinal properties:

Suppose $\varphi$ is downwards absolute between end extensions of $T_0$, and $T_0\subseteq T_1$. Then $\varphi$ is downwards absolute between end extensions of $T_1$.

You're never going to get more "potential variety" by adding axioms, regardless of whether they ramp up the consistency strength of the theory. The crucial point here, and what I think should be the real takeaway of this answer, is:

$$\mbox{A large cardinal axiom is more than just a strong consistency principle.}$$

EDIT: another important takeaway, relating to your comment below (and I think worth incorporating into this answer for completeness), is:

$$\mbox{Forcing is not the only tool for producing models.}$$

If $T$ is a consistent theory extending ZFC, there is no reason to believe that models of $T$ can always be produced from arbitrary models of ZFC by forcing. While forcing is a powerful tool (indeed, one of the very few we have) for proving relative consistency results, the inability of forcing to produce a model of a given theory from an arbitrary model of ZFC is in no way evidence that that theory is inconsistent.

Let me end by giving an explicit example of a theory with the property you describe in the latter part of your question:

A theory $T$ (for simplicity: extending ZFC) has property $(+)$ if PRA proves "Con($T$ + there is an inaccessible) iff Con($T$ + there are no inaccessibles)."

Of course there are trivial examples (ZFC + $0\not=0$). To construct a nontrivial example of $(+)$, the fastest thing to do is to turn to inner models:

Let $T$ be the theory ZFC + "there is an $L$-inaccessible."

The point is that since $L$ is definable $T$ makes sense, and since $L$ is invariant under end extensions (precisely: if $M\subseteq_{end}N$ then $L^N=L^M$) $T$ is preserved by taking arbitrary inner models and arbitrary forcing extensions. In particular, if $M$ is a countable model of $T$, then:

  • $L^M$ is a model of $T$ + "There is an inaccessible cardinal."

  • $M_\infty[G]$ is a model of $T$ + "There is no inaccessible cardinal," where:

    • if $M$ has no inaccessible cardinals, then $M_\infty=M$ and $G=\emptyset$ (so $M_\infty[G]=M$);

    • if $M$ has a unique inaccessible cardinal $\kappa$, then $M_\infty=M$ and $G$ is $Col(\omega,\kappa)^M$-generic over $M$; and

    • if $M$ has at least two inaccessible cardinals with $\kappa_1, \kappa_2$ being the first and second inaccessible cardinals respectively, then $M_\infty=(V_{\kappa_2})^M$ and $G$ is $Col(\omega,\kappa_1)^{M_\infty}$-generic over $M_\infty$ (and we can equivalently replace these last two "$M_\infty$"s with "$M$"s if we want).

All of the above is argued semantically, but can be turned into a PRA-proof by the usual tricks. So $T$ does in fact have property $(+)$.

However, the broader point above stands: since $T$ is an extension of ZFC, we can't force a non-inaccessible to become inaccessible over a model of $T$.

  • $\begingroup$ I think that (*) (and your reformulation of it) should begin with "Assuming ZFC + Con(ZFC) is consistent, ...." I don't see how to get this result assuming only the consistency of ZFC. $\endgroup$ May 14, 2018 at 21:33
  • $\begingroup$ @AndreasBlass A fair point, I slipped up there. Fixed! $\endgroup$ May 14, 2018 at 22:29
  • $\begingroup$ @NoahSchweber: Thanks for this very interesting answer! Would I be incorrect in inferring from this answer that there can be no models of $ZFC$ + "There exists an inaccessible cardinal"? If this inference would, in fact, be incorrect, what assumptions would be necessary to infer the existence of a model of $ZFC$ + "there exists an inaccessible cardinal (other than assumptions of inconsistency)? $\endgroup$ May 31, 2018 at 16:00
  • $\begingroup$ @ThomasBenjamin "Would I be incorrect in inferring from this answer that there can be no models of ZFC + "There exists an inaccessible cardinal"?" Are you sure you meant what you intended? The answer is absolutely no - the key point being that forcing is not the only way to build models. For your last sentence, I think your "existence" should be "nonexistence," and the answer is: precisely the inconsistency of ZFC + "there exists an inaccessible cardinal." This is an instance of the completeness theorem. $\endgroup$ May 31, 2018 at 16:31
  • $\begingroup$ Precisely: if ZFC + "There is an inaccessible cardinal" is consistent, then it has some model $M$; however, this model is not a forcing extension of any model of ZFC + "There is no inaccessible cardinal." There's no tension here, since in general why should we expect a model of a theory $T_0$ to automatically have to be a forcing extension of a model of a different theory $T_1$? $\endgroup$ May 31, 2018 at 16:34

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