0
$\begingroup$

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.)

$\endgroup$
  • 1
    $\begingroup$ In which set theory are you working? ZF? NF? Something else? $\endgroup$ – Asaf Karagila Sep 4 '12 at 18:25
  • $\begingroup$ @AsafKaragila: See edit. $\endgroup$ – Anton Golov Sep 4 '12 at 18:27
  • 1
    $\begingroup$ 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 $\endgroup$ – Trevor Wilson Sep 4 '12 at 18:40
8
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ The set I'm asking about has one extra restriction: the $V$ in question is the given set, i.e. $V = V^V$. $\endgroup$ – Anton Golov Sep 4 '12 at 18:12
  • $\begingroup$ @AntonGolov I don't understand. So $V$ is a singleton? Otherwise $V$ and $V^V$ have different cardinalities so cannot be equal. $\endgroup$ – Alex Becker Sep 4 '12 at 18:15
  • $\begingroup$ Good point. I guess that gives me my set then, heh. I was hoping for something bigger, didn't think about the cardinalities issue. :) $\endgroup$ – Anton Golov Sep 4 '12 at 18:18
  • $\begingroup$ @AntonGolov There's also the issue of whether you mean equal or isomorphic. $\endgroup$ – Alex Becker Sep 4 '12 at 18:20
3
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Anton Golov Sep 4 '12 at 18:19
  • $\begingroup$ 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). $\endgroup$ – Anton Golov Sep 4 '12 at 18:30
  • $\begingroup$ 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. $\endgroup$ – André Nicolas Sep 4 '12 at 18:35
  • $\begingroup$ @HenningMakholm: Thanks, fixed. $\endgroup$ – André Nicolas Sep 4 '12 at 19:05
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.