TITLE: Intuitive descriptions of some large cardinals
QUESTION [6 upvotes]: 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.

REPLY [7 votes]: (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.