# What sets satisfy $V = V^V$?

I'm not sure if said set exist or whether it is unique, but what name could I use to find more about it and what kind of interesting properties does it have?

Clarification edit: I meant a set $V$ such that $V = V^V$. I am curious about the concept and would like to read more about it; I don't mind which formalism this is in. (From the answers, I understand that this isn't possible in most.)

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In which set theory are you working? ZF? NF? Something else? – Asaf Karagila Sep 4 '12 at 18:25
@AsafKaragila: See edit. – Anton Golov Sep 4 '12 at 18:27
If you like applying functions to themselves, you should check out something like this paper on finite left-distributive algebras and embedding algebras: arxiv.org/abs/math/9209202v1 – Trevor Wilson Sep 4 '12 at 18:40

Edit: Apparently you want to know the name for the set $V$ which satisfies $V=V^V$. I see no reason to believe a priori that such a set exists or is unique. However, we can argue by cardinality that $|V|=1$ if $V$ exists. Then it becomes an issue of whether you mean $V=V^V$ or $V\cong V^V$. The first is impossible, since the function $f:v\to v$ which maps the single element $x$ to itself is not literally the same as $x$. The second applies to all singletons.

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The set I'm asking about has one extra restriction: the $V$ in question is the given set, i.e. $V = V^V$. – Anton Golov Sep 4 '12 at 18:12
@AntonGolov I don't understand. So $V$ is a singleton? Otherwise $V$ and $V^V$ have different cardinalities so cannot be equal. – Alex Becker Sep 4 '12 at 18:15
Good point. I guess that gives me my set then, heh. I was hoping for something bigger, didn't think about the cardinalities issue. :) – Anton Golov Sep 4 '12 at 18:18
@AntonGolov There's also the issue of whether you mean equal or isomorphic. – Alex Becker Sep 4 '12 at 18:20

Probably not what you intended, but since you explicitly don't mind which formalism:

In set theory with Aczel's anti-foundation axiom, there is exactly one set $x$ such that $x=\{\langle x,x\rangle\}$, and its singleton $\{x\}$ then satisfies your condition.

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If by $V$ you mean "the von Neumann universe," that is, the class of all sets, then $V$ is not a set, let alone $V^V$.

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Could you clarify what you mean with "the universe"? I simply meant a set $V$ such that $V = V^V$; Alex has already pointed out that there's only one such set, though. – Anton Golov Sep 4 '12 at 18:19
Ah, thanks, will keep in mind. I'm used to $V$ being used for arbitrary sets, but that's probably due to my language of education (Dutch). – Anton Golov Sep 4 '12 at 18:30
In English, $V$ would be an unusual name for a set. It has become standard notation for "the universe," as in the Axiom of Constructibility $V=L$. By the way, in Simon Stevin's opinion, Dutch was the language most suitable for mathematics. – André Nicolas Sep 4 '12 at 18:35
@HenningMakholm: Thanks, fixed. – André Nicolas Sep 4 '12 at 19:05

Such a set does not exist.

We must have $|X| = 0$ or $X = |1|$, or else $X^X$ has more elements than $X$. If $X = 1$, say $X = \{x\}$, then $x = (x,x)$, which is impossible. If $X = 0$ this is also impossible, because $\emptyset \in \emptyset^\emptyset$ as Arthur pointed out.

Here is a simpler proof not involving Cantor's theorem. If $X = X^X$ and $X$ is nonempty them by the axiom of regularity $X$ has an element $x$ with minimal rank. However, $x$ is a (nonempty) function $X \to X$, so it contains some ordered pair $(y,z)$ of elements of $X$, and then $y$ and $z$ have smaller rank than $x$, a contradiction.

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I think there is exactly one function $\emptyset \to \emptyset$. This function is usually called the empty set, and so $\emptyset^\emptyset = \{ \emptyset \} \neq \emptyset$. – arjafi Sep 4 '12 at 18:27
Good catch, thanks. – Trevor Wilson Sep 4 '12 at 18:32
@Arthur: I usually hear it called the empty function. – Hurkyl Sep 4 '12 at 18:43