# Intuitive descriptions of some large cardinals

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.

• Round about here books.google.com/… Rudy Rucker in his Infinity and the Mind attempts to give intuitive descriptions of large cardinals. – Todd Trimble Jun 25 '16 at 22:36
• 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. – Anindya Jun 27 '16 at 3:30
• Large cardinals are not intuitive. – Franz Lemmermeyer Jun 28 '16 at 4:50

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.