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Many years ago in an oral exam I was asked, what could be concluded from the existence of an inaccessible cardinal in ZFC? I knew that would provide a model for ZFC and imply the consistency of ZFC.

Then I was asked, what could be concluded from the existence of two inaccessible cardinals? I had no clue then and it still bothers me to this day. I've never seen anything like that in the literature.

Any insights on what can be inferred from the existence of two inaccessible cardinals?

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    $\begingroup$ ZFC plus there exists an inaccessible cardinal is consistent? $\endgroup$ – André Nicolas Jul 18 '16 at 4:03
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    $\begingroup$ @AndréNicolas is right. Cutting off the cumulative hierarchy at the second inaccessible cardinal gives you a natural modelof "ZFC + there exists an inaccessible cardinal." Starting with that model, you could apply the Löwenheim-Skolem theorem and transitive collapsing to get transitive models of "ZFC + there exists an inaccessible cardinal" of all infinite cardinalities below the second inaccessible cardinal. $\endgroup$ – Andreas Blass Jul 18 '16 at 5:04
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    $\begingroup$ The consistency of "all projective sets are Lebesgue measurable". $\endgroup$ – Asaf Karagila Jul 18 '16 at 5:51
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    $\begingroup$ What a pleasant surprise to see a comment by @AndreasBlass. I don't recall if you had asked the question or Peter Hinman. :-) In hindsight, "ZFC + there exists an inaccessible cardinal" seems so obvious. The wikipedia entry on large cardinals also mentions that point under "Hierarchy of consistency strength". I was expecting something more exotic like Asaf's response. I'll have to investigate that one. Thanks! $\endgroup$ – Gary Kibble Jul 18 '16 at 11:35
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The obvious answer is that you get the consistency of $\sf ZFC$ with an inaccessible. But you get more. You get the existence of a transitive model of this theory. But you actually get more. Since an inaccessible cardinal is the limit of worldly cardinals,1 you get a transitive model of $\sf ZFC$+There exists an inaccessible cardinals+There is a proper class of worldly cardinals.

But of course, this is the dry an obvious answer. Let's take a look at a few theories which have been proved to be equiconsistent with the existence of an inaccessible cardinal. Because since you get a model of $\sf ZFC$+There exists an inaccessible cardinal, we actually get a model of these theories.

  1. $\sf ZF+DC$+Every set of reals is Lebesgue measurable; or $\sf ZFC$+Every projective set of reals is Lebesgue measurable. Shelah proved that it is enough to consider $\Sigma^1_3$ sets of reals in order to conclude an inaccessible must exist. Similarly, you can replace Lebesgue measurable by "every uncountable set of reals contains a perfect subset" (which holds in Solovay's model), which implies $\omega_1$ is a limit in $L$, so in the presence of $\sf DC$, implies the existence of an inaccessible cardinal in an inner model.

  2. The consistency of $\sf ZF$+There exists an infinite Dedekind-finite set+Every two Dedekind-finite cardinals are comparable. This was proved by Sageev, the inaccessible is most likely unnecessary, but we don't quite know yet.

  3. The consistency of the Kelley-Morse set theory; but also the fact that $\sf ZFC_2$ has at least two non-isomorphic models (one without inaccessible cardinals, and one with exactly one).

  4. The consistency of $\sf ZFC$+There are no Kurepa trees.

  5. The consistency of $\sf ZF$+For every $\alpha$ there is a set $X_\alpha$ which is a countable union of countable sets, but $\mathcal P(X_\alpha)$ can be mapped onto $\omega_\alpha$. Like Sageev's result, it is unclear if the inaccessible cardinal is actually needed (and unlike Sageev's result, this one was never published as a paper: it was announced in Notices of the AMS, and it should appear in a Ph.D. thesis from ages ago).

  6. The consistency of $\sf ZFC$+$\omega_1$ is inaccessible for reals. There are various results in descriptive set theory about $\sigma$-ideals on Polish spaces which have implications when assuming $\omega_1$ is inaccessible to reals, which means that these consequences become true in some models.

This list is by no means complete, or even remotely close to being complete. But it does give you a small taste of what people did with just inaccessible cardinals over the years. Frankly, however, for a lot of interesting results inaccessible cardinals turn out to be too weak and too mundane to provide us with the necessary structure for carrying out these results.


Footnotes.

(1) $\kappa$ is worldly if $V_\kappa$ is a model of $\sf ZFC$. Every inaccessible cardinal is worldly, but the least worldly has countable cofinality, and an inaccessible cardinal has a club of worldly cardinals below it.2

(2) Note that if we replace the Replacement schema by its second-order axiom, then we do get that $V_\kappa\models\sf ZFC_2$ if and only if $\kappa$ is inaccessible.

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  • $\begingroup$ Excellent answer. Much appreciated! $\endgroup$ – Gary Kibble Jul 18 '16 at 19:07
  • $\begingroup$ @Asafkaragila: Regarding 3, does it matter that in $ZFC^2$, the model of $ZFC^2$ with no inaccessible cardinals is an initial segment of the model of $ZFC^2$ with exactly one inaccessible cardinal with regards to the models being non-isomorphic? Does the consistency of $ZFC$ +"There exists two inaccessible cardinals" prove the consistency of both $ZFC$ + "There are no inaccessible cardinals" and $ZFC$ + "There exists one inaccessible cardinal"? $\endgroup$ – Thomas Benjamin Jul 31 '18 at 16:58

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