I was trying to formulate intuitive descriptions of some large cardinals.
Roughly something equivalent to "A manifold is an object which looks like patches of $R^n$ glued together". Not perfectly rigorous, but hopefully conveys the basic picture.
Here are three descriptions I have in mind:
1) An inaccessible cardinal is a set so large that it can't be reached from smaller infinite sets using unions and power set operations.
2) A measurable cardinal is a set so large that it can't be reached from smaller infinite sets using any set theoretic formula. (I am going for V $\neq$ L)
3) A Reinhardt cardinal is a set so large that all possible properties of the entire set theoretic universe are also true of this set. (If I am correct, this would capture the intuition of why Reinhardt cardinals don't exist. They are simply too ambitious).
I am pretty confident that description 1 is correct in essentials, but are 2) and 3) hopelessly off ? If so, how would one appropriately modify the descriptions ? Or are object like measurable cardinals just too abstract to be expressed in anything but technical definitions ?
Apologies in advance if this question is too elementary for MathOverflow.

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    $\begingroup$ Round about here books.google.com/… Rudy Rucker in his Infinity and the Mind attempts to give intuitive descriptions of large cardinals. $\endgroup$
    – Todd Trimble
    Jun 25, 2016 at 22:36
  • 1
    $\begingroup$ Todd, Rudy Rucker's book was what got me wondering about this question. He stops short of explaining measurable cardinals, except for saying "they are so big that comparing them to run-of-the-mill inaccessibles is like comparing $\omega$ to 2". But the definition of a measurable cardinal does little to convey WHY it is so large. $\endgroup$
    – Anindya
    Jun 27, 2016 at 3:30
  • $\begingroup$ Large cardinals are not intuitive. $\endgroup$ Jun 28, 2016 at 4:50

1 Answer 1


(1) seems okay, but I'm afraid that most large cardinal properties beyond inaccessibility probably aren't going to admit such simple formulations.

I'm not sure I understand the precise meaning of (2), but in any case it doesn't seem right. For one thing, the existence of $0^\sharp$ is weaker than the existence of a measurable cardinal and still implies $V \ne L$. Also if $0^\sharp$ exists then there are many cardinals, namely indiscernibles for $L$, that seem to satisfy (2) but are not measurable.

I don't think (3) is right either. To me it sounds like it describes a cardinal $\kappa$ such that $V_\kappa$ is an elementary substructure of $V$. The existence of such a cardinal is much weaker than the existence of a Reinhardt cardinal, and in fact is equiconsistent with $\mathsf{ZFC}$ by a compactness argument.

  • $\begingroup$ Thanks, Trevor. I was just wondering if there is any intuitive way to convey WHY one expects a Reinhardt cardinal to be humongously large - to the point of inconsistency with ZFC. $\endgroup$
    – Anindya
    Jun 27, 2016 at 3:27
  • $\begingroup$ Reinhardt cardinals are inconsistent with ZFC by Kunen's inconsistency theorem and Hugh Woodin has conjectured them to be inconsistent with ZF as well. I don't know any intuitive explanation of either of these things. $\endgroup$ Jun 29, 2016 at 19:27
  • $\begingroup$ Why Reinhardt cardinals would be considered very large is a simpler matter. Consider that (to use some common examples of large cardinals) every measurable cardinal is a limit of inaccessible cardinals, every Woodin cardinal is a limit of measurable cardinals, every superstrong cardinal is a limit of Woodin cardinals, etc., and Reinhardt cardinals occur very far up in this "hierarchy" of large cardinals. $\endgroup$ Jun 29, 2016 at 19:31

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