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.

4$\begingroup$ 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 runofthemill inaccessibles is like comparing $\omega$ to 2". But the definition of a measurable cardinal does little to convey WHY it is so large. $\endgroup$– AnindyaJun 27, 2016 at 3:30

$\begingroup$ Large cardinals are not intuitive. $\endgroup$– Franz LemmermeyerJun 28, 2016 at 4:50
1 Answer
(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. $\endgroup$– AnindyaJun 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