I'm having some difficulty making this question precise, so bear with me. I have read that some "small objects" can have consequences on the larger set theoretic universe, such as showing that certain kinds of large cardinals exist, but I don't understand how this process works.

I can see how you can have constructions on the natural numbers that have high consistency strength (i.e. a subset of $\Bbb N$ that is a countable model of ZFC), but can this process be "reversed" to actually get the large cardinal? For example, I might know that ZFC + an inaccessible cardinal is consistent given a countable model of such, but I don't think that implies that $V$ contains an inaccessible cardinal.

I often see $0^\#$ discussed in these terms, but I'm utterly baffled by that definition and it's not clear to me whether you can actually extract large cardinals from it.

  • $\begingroup$ Well a proof of the existence of large cardinals (though it would also prove the inconsistency of ZFC...) is essentially a natural number so in a sense, yes... $\endgroup$ – Max Dec 21 '17 at 17:12
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    $\begingroup$ To address your embedded question "but I don't think that implies that V contains an inaccessible cardinal." You don't -- as you suspected -- get back an inaccessible from a (countable) transitive model. In fact, you don't even get back in inaccessible in an inner model. This is actually a fairly easy exercise. $\endgroup$ – Stefan Mesken Dec 21 '17 at 17:46
  • $\begingroup$ A link that is 100% mathjax is less useful than one that can be "opened in new tab". I might replace that with $0^{\#}$ (zero sharp) (whatever the URL is that I can't see without leaving this page, so losing this comment). $\endgroup$ – Eric Towers Dec 21 '17 at 23:02

No, you do not get large cardinals in the actual universe out of the existence of such small sets. The reason is that if $\kappa$ is least such that $V_\kappa$ is a model of set theory, then $V_\kappa$ also contains all these small sets, but has no large cardinals (by minimality of $\kappa$).

Instead, what we get is the consistency of extensions of set theory with large cardinals. In practice we actually tend to get more, namely, that the large cardinals exist in certain inner models.

For instance, $0^\sharp$ can be defined as a certain set of numbers. If it exists, then in $L$ there are many inaccessible cardinals, although you cannot prove that it gives you any large cardinals in $V$.

For another example, to see how flexible the format is, consider the statement that all projective sets of reals are Lebesgue measurable. This is a statement about the first order theory of the reals (and it can be stated as a schema of second order arithmetic, via appropriate coding). You cannot get large cardinals from it directly, but it implies that $\omega_1^V$ is an inaccessible cardinal in $L$.

  • $\begingroup$ So doesn't this mean that if the background theory is $ZFC+V=L$, then we do get inaccessible cardinals from small sets? Or does this not work because $0^\sharp$ is inconsistent with $V=L$? $\endgroup$ – Mario Carneiro Dec 21 '17 at 18:29
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    $\begingroup$ No, you still don't, thanks to the argument in the first paragraph of the answer. For the cases we find in practice, such as the two examples I give, the assumptions readily imply that $V\ne L$. There are some curious additional examples, considered by H. Friedman, where the statements we have are compatible with (and independent of) $V=L$. Still, you do not obtain large cardinals from the statements, only their consistency, although now you cannot even follow the usual route to establish this, since they do not imply large cardinals in inner models. $\endgroup$ – Andrés E. Caicedo Dec 21 '17 at 19:07
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    $\begingroup$ A reference is MR1670057 (2002b:03108). Friedman, Harvey M. Finite functions and the necessary use of large cardinals. Ann. of Math. (2) 148 (1998), no. 3, 803–893. Harvey has many additional examples of a similar nature in notes available from his page. $\endgroup$ – Andrés E. Caicedo Dec 21 '17 at 19:09

As phrased, it's not entirely clear what you are asking. Several possible answers, that don't have all that much in common, come to mind:

Certain structures, which can be coded as subsets of $\omega$, do imply that there are inner models with large cardinals. E.g. $0^\#$ implies that $L$ has a proper class of inaccessibles, $0^\dagger$ (read "zero dagger") implies that there is an inner model with a measurable cardinal, "zero pistol" (whose symbol is apparently not supported by MathJaX) implies that there is an inner model with a strong cardinal, ... They don't, however, imply that these large cardinals exist in our background universe.

Since there is no formal definition of 'large cardinal', the following has to be taken with a grain of salt: Subsets of $\omega$ don't provably add large cardinals to our background universe.

Why? Well, suppose $A \subseteq \omega$ did add/witness a large cardinal in $V$. Let $\kappa$ be the least wordly cardinal of $V$ (which should exist because $A$ adds a large cardinal). Then $A \in V_{\omega +1} \subseteq V_{\kappa}$ and $(V_\kappa; \in) \models \mathrm{ZFC}$ but the large cardinal that $A$ witnesses doesn't exist in $V_\kappa$.

Anything can be added by a subset of $\omega$:

(Jensen.) There is always a class generic $G$ and some $x \subseteq \omega$ in $V[G]$ such that $$ V[G] = L[x] \supseteq V $$ So any large cardinal that exists in $V$ is, in a sense, added by the real $x$.

My hope is that this helps you to get to the essence of your intuitively justified question which would then allow us to think about a more precise answer.

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    $\begingroup$ I think your formulation of Jensen's theorem is unnecessarily confusing. Perhaps simply saying that $V[G]=L[x]$ is better, since this makes it clear that it is $V[G]$ rather than $V$ itself that is coded by the real. The reason why this matters is that the large cardinal itself may be destroyed by adding $x$, and anyway some large cardinals imply that the universe is not $L[X]$ for any set $X$. $\endgroup$ – Andrés E. Caicedo Dec 21 '17 at 17:48
  • $\begingroup$ @Andrés I agree. I didn't realize, while typing, that this is indeed misleading. I'll fix it right away. $\endgroup$ – Stefan Mesken Dec 21 '17 at 17:52

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