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I want to understand the viewpoint that existence of canonical inner model for a large cardinal notion is strong evidence for its consistency. For example, below is Trevor Wilson's answer to What "forces" us to accept large cardinal axioms?:

Second, there is "fine structure" which gives canonical models for the smaller large cardinal axioms (so far, up to Woodin cardinals and a bit further.) It seems reasonable to expect that a systematic study of the structure of the models of a theory would eventually reveal the inconsistency of the theory if it were inconsistent, and this has not happened yet.

Andrés E. Caicedo's answer to Arguments against large cardinals:

The point of the inner model program (and of its most recent offspring, descriptive inner model theory) is to develop fine structural ("$L$ like") models for large cardinals. These models are canonical in several precise ways, and have a rich internal structure that many set theorists take as evidence of the consistency of the large cardinals under consideration.

Stefan Geschke's answer to the same question:

There is the so-called inner model program where one assumes the existence of a certain large cardinal and tries to build an (easily controllable) smallest model of set theory in which there is such a large cardinal and which contains all the ordinals. The idea is that because we have a good understanding of the final inner model, we would notice during the construction of the model if there were any problems with the consistency of the large cardinal in question.

Can someone elaborate on why the rich structure of the canonical inner model provides evidence for consistency of the corresponding large cardinal? Below are some more precise questions:

  1. Say $U$ is a $\kappa$-complete normal measure on $\kappa$. What exactly does "fine structure" of the model $L[U]$ refer to? I am quite ignorant in inner model theory (maybe I will be more motivated to learn it if I get a satisfactory answer) so below are just words that I've seen here and there: the core model $K$ is the union of all mice, and $L[U]$ exists iff there is a nontrivial elementary embedding from $K$ into itself, and this gives a definition of $L[U]$ that "approximates from below", in contrast to the definition using relative constructibility. Is this what "fine structure" means here?

  2. But then we need to believe in mice in the first place. For example, how to believe in the existence/consistency of $0^\sharp$? Of course it follows from analytic determinacy, but I don't know enough examples of analytic sets to convince myself that analytic determinacy is obviously true...Is there any other good reason?

  3. Can we say the fine structure of $L$ is a strong evidence for consistency of ZFC?

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    $\begingroup$ It may not answer your question, but the fact that Gödel's $L$ is frequently used to provide the ordinal analysis for extensions of $\mathsf{KP}$ may give a hint: The way to do the ordinal analysis for set theory is constructing a term model modulo a canonical system of ordinals. There is no way to construct a term model in general, but it is possible if we rely on $L$ in which every set is canonical. We may understand the inner model theory program as a project providing a term model for large cardinal axioms modulo some minimal notions (like, ordinals.) $\endgroup$
    – Hanul Jeon
    Commented Apr 14 at 18:40
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    $\begingroup$ Regarding $0^\#$, you may need to learn more about analytic sets to have an informed opinion. One nice result (due to Steel) is that every real has a sharp if and only if every analytic set is either Borel or analytic complete. Again the only known proof of this result uses some serious inner model theory, namely Jensen's covering lemma via the proof of Harrington's theorem on the equivalence of analytic determinacy and sharps. $\endgroup$
    – Gabe Goldberg
    Commented Apr 16 at 16:06
  • $\begingroup$ @GabeGoldberg Sorry for digging this up---is there an easy example in $L$ of a non-complete true analytic set? And is the result wholly due to Steel? "Analytic completeness implies sharp" is Theorem 4.3 in Harrington's analytic determinacy paper, where he attributes the converse to Steel. $\endgroup$
    – n901
    Commented Sep 19 at 23:52
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    $\begingroup$ @n901 I think you're right about attribution. Doesn't Harrington's proof show that the set of $b\in L$ such that some $a \leq_T b$ codes an end extension of $L_{\omega_1(b)}$ is $\Sigma^1_1$ but not complete in $L$? $\endgroup$
    – Gabe Goldberg
    Commented Sep 20 at 14:39

2 Answers 2

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The explanation is philosophical rather than mathematical.

The idea is simply that the inner-model theory provides a rich account of what it would be like for the large cardinal axioms to be true, and this account is so detailed and full that it gives us what seems to be a deep understanding of the large cardinal hypothesis in question, that we might expect to be confident that if there were any problematic issue about the hypothesis, it would have been revealed during the analysis.

By building the inner models and elucidating the accompanying fine structure, we gain an understanding of the hypotheses behind it in the same way that an engineer gains insight to the workings of an engine by taking it completely apart and reassembling it, like my father and older brothers showed me in the family shop when I was a boy. The inner model theorists have taken the universe with a measurable cardinal apart and reassembled it with fine structure. They really know what it makes it hum.

I believe it is essentially this perspective that is expressed by the quotations you have made in the original question.

Nevertheless, one must keep in mind that this view expresses a philosophical expectation rather rather than a mathematical certainty. And indeed, we know in a deep way that no fine-structural account of a theory can imply the consistency of the theory as a general mathematical theorem. For example, if ZFC is consistent, then we know that it is consistent that ZFC + V=L + ¬Con(ZFC), with the point being that in any world fulfilling this theory, we would have the full fine-structural account of $L$, and yet still think that ZFC is inconsistent. In this world, the set theorists will have taken the world apart and put it together again by means of the constructible hierarchy, and yet still the theory is inconsistent here. So we cannot deduce in general from the fine-structural account of $L$ that ZFC is consistent.

The same goes for all the large cardinal theories. It is consistent to think that ZFC plus $V=L[\mu]$ for a measure $\mu$ on a measurable cardinal $\kappa$, and yet also think that the theory "ZFC+ there is a measurable cardinal" is inconsistent. Such a model betrays the philosophical idea that fine-structural accounts lead to consistency. The same goes for any theory extending ZFC.

Nevertheless, many set theorists find those kinds of theories, where we assert a theory $T$ but also assert $\neg\text{Con}(T)$, to be deeply flawed in some fundamental way, even though they may be consistent. There are philosophical reasons to find them incoherent. And the success of those philosophical arguments is related to the success of the philosophical arguments about which you inquire.

Meanwhile, the point of the rebuttal is not to advocate for $T+\neg\text{Con}(T)$, but rather to point out merely that one cannot deduce consistency from fine structure.

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    $\begingroup$ There is this thing in number theory called Siegel zeros, which are counterexamples to the generalized Riemann hypothesis. Number theorists study them in great details so that they can prove theorems by cases: if GRH is true, use all the nice consequences; if GRH is false, use all the things we know about the strange world of Siegel zeros...By the same logic, I guess we should believe in the independence of GRH :) $\endgroup$
    – n901
    Commented Apr 15 at 0:36
  • $\begingroup$ That is interesting, although I am unsure whether there is the same thoroughness of the accounts in the GRH case as in the inner models. The inner model theory seems aimed at providing something like a complete picture of the nature of the inner model in which the large cardinal hypothesis holds. Of course, this isn't fully true, for the reasons I've mentioned in my answer, but it seems more true than is the case for the Siegel zeros. $\endgroup$
    – Joel David Hamkins
    Commented Apr 15 at 0:58
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    $\begingroup$ @n901 I think that if you apply the "same logic" to GRH, then the conclusion is that the superior mathematical coherence of the "GRH is true" universe compared to "Siegel zeros exist" universe is considered to be evidence that GRH is true, even though we can't prove GRH. Scott Aaronson has made a similar case for the truth of $\mathsf{P}\ne\mathsf{NP}$; see this MO question. $\endgroup$
    – Timothy Chow
    Commented Apr 15 at 12:12
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    $\begingroup$ @TimothyChow Terry Tao discussed exactly about Siegel zeros in his answer to that question, and he doesn't seem to claim the "GRH is true" universe is superior...Just that the methods used in two universes are different. In fact there are theorems easier to prove in the Siegel zero universe (twin prime conjecture). Anyway my intent was to argue that this logic should be applied with great caution. $\endgroup$
    – n901
    Commented Apr 15 at 13:15
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    $\begingroup$ @n901 If the GRH universe is not superior, then perhaps that should be regarded as evidence against GRH. Anyway, of course mathematicians are congenitally cautious about anything that is not proved. That doesn't mean there is no such thing as evidence for and against a conjecture. In every arena of mathematics, if we deduce increasingly elaborate consequences of a conjecture that all seem to hang together and make sense, we tend to regard that as evidence for the conjecture. Indeed, in the natural sciences, falsifiable statements that aren't falsified are all we have to go on! $\endgroup$
    – Timothy Chow
    Commented Apr 15 at 14:56
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There is a second way inner model theory argues for the consistency of large cardinals. This is by enabling us to prove consistency strength lower bounds. If you don’t believe that one can construct a Lebesgue nonmeasurable set using countable choice, then you must believe the consistency of ZFC. If you don’t believe the singular cardinals hypothesis is provable in ZFC, then you must believe the consistency of a measurable cardinal. So one’s confidence in seemingly unrelated mathematical facts can be transferred to a confidence in various large cardinal hypotheses.

Results like this do not exist at the level of supercompact cardinals because inner model theory does not yet reach that high. Therefore for all we know it may be that much of the structure we see in forcing theory at the level of supercompacts (eg Martin’s maximum) is an illusion because we are using the wrong (perhaps inconsistent!) hypotheses. There can be no corresponding anxiety at the level of Woodin cardinals thanks to inner model theory.

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  • $\begingroup$ Thanks. I'm aware of some of these results, and I find them more convincing than the fine structure argument. Is countable choice necessary for Shelah's theorem on nonmeasurable set? Shelah's original paper only used the regularity of $\omega_1$, but I am really asking whether ZF suffices. Without countable choice we don't have a nice $\sigma$-algebra to do measure theory, but it should still make sense to consider Borel-codable sets, their images under continuous maps, etc., and ask if they are measurable? $\endgroup$
    – n901
    Commented Apr 16 at 3:10
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    $\begingroup$ ZF does not suffice. Elliot Glazer has some nice results about this, though I’m not sure if they’re published yet. I believe second order number theory plus the countable choice scheme plus a scheme asserting that every projective set is Lebesgue measurable is equiconsistent with ZFC; I don’t actually know that this is true but suspect it is… $\endgroup$
    – Gabe Goldberg
    Commented Apr 16 at 12:36
  • $\begingroup$ It looks like one direction (from ZFC to second-order arithmetic plus $\mathsf{AC}_\omega$ plus every projective set is Lebesgue measurable) should follow from collapsing every ordinal to a countable ordinal and considering $L(\mathbb{R})$ over the extension. (The collapse forcing is pretame, so the forcing extension satisfies $\mathsf{ZFC}^-$. $\mathbb{R}$ becomes a proper class over the forcing extension, but we should be able to define $L(\mathbb{R})$ in a way to define $L(\mathrm{Ord}^\omega)$ over $\mathsf{ZFC}$.) Is that what you thought? $\endgroup$
    – Hanul Jeon
    Commented Apr 18 at 23:19
  • $\begingroup$ @HanulJeon Yes, that's basically what I was thinking. I think it is not necessary to look at $L(\mathbb R)$ over the extension, you just collapse everything and then the sets of numbers you wind up with give you your model of second-order arithmetic. The other direction I'm not so sure about since it's been a long time since I looked at Shelah's lower bound proof, but it should just be possible to build $L$ in your model of second-order arithmetic and show ZFC holds there. $\endgroup$
    – Gabe Goldberg
    Commented Apr 23 at 16:38
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    $\begingroup$ @HanulJeon $Z_2 + PM$ (projective measurability) is equiconsistent with Z, and $Z_2 + PM + AC_{\omega}$ with ZF. The reverse implications are basically by your argument, noting $Con(Z) \Leftrightarrow Con(ZC + KP + V=L_{\omega_{\omega}}).$ For the forward implications, we need to show $L \models Z$ in $Z_2 + PM.$ All but the axiom of power set holds in $L$ just using $Z_2.$ If power set fails in $L,$ we can define a (class) $\omega_1$-sequence of reals and use Raisonnier's construction of a rapid filter to definable a class of reals with inner measure 0 and outer measure 1. $\endgroup$
    – Elliot Glazer
    Commented Nov 19 at 20:56

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