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Looking at the ZFC axioms (in the Wikipedia version), the axiom of infinity stands out because it contains an extremely specific construction. This seems rather unelegant to me, therefore I've thought about whether it could be replaced by a more generic version.

Now one property of the set natural numbers in standard construction (and indeed, of any limit ordinal, if I understand it correctly) is that it is the union of all its elements.

Now from trying it seems to me (although I have no formal proof of that) that the only finite set with this property is the empty set (for which it trivially holds), because you cannot construct a non-empty finite set which has that property without violating the axiom of foundation.

Therefore I thought of the following

Axiom of Non-Constructive Infinity: There exists a non-empty set which is the union of all its members.

The question is: Is ZFC with Infinity replaced by Non-Constructive Infinity equivalent to ZFC? (And what about ZF without C?)

It is easy to prove from the axiom of foundation that any non-empty set which is the union of all its members must contain the empty set (because all its members are subsets, and thus any non-empty member cannot be disjunct with it). However I neither managed to prove that such a set must for each element $x$ also contain $x\cup\{x\}$, nor was I able to construct a counterexample. (But then, even if a counterexample existed, it might still be possible to construct such a "standard-infinity set" from any "self-union set", so it would by itself not negatively answer my question).

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Note that we have $A=\bigcup A$ if $A=\omega\cup\{\{42\}\}$, but the successor of $\{42\}$ is not $\in A$. –  Hagen von Eitzen Apr 21 '14 at 20:19
Hmmm ... with $A=\omega\cup\{\{42\}\}$ I get $\bigcup A=\omega\cup\{42\}\ne A$. –  celtschk Apr 21 '14 at 20:39

2 Answers 2

up vote 5 down vote accepted

I claim that is suffices to show that such set is infinite. We can prove that there are only finitely many sets of a given finite rank, and they are all finite. So an infinite set must have an infinite rank, so $\omega$ must exist, which is an inductive set.

Let us that if $x$ a non-empty set of a finite rank then $\bigcup x\neq x$. $\DeclareMathOperator{rank}{rank}$

It suffices to show that $\rank(\bigcup x)<\rank(x)$, when $x$ is non-empty and has a finite rank.

Recall that for finite sets it holds that $\rank(x)=\max\{\rank(y)+1\mid y\in x\}$. Therefore if $a\in\bigcup x$ then there is some $b\in x$ such that $a\in b$, so $\rank(a)+1<\rank(x)$.

Therefore $\max\{\rank(a)+1\mid a\in\bigcup x\}=\rank(\bigcup x)<\rank x$.

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"We can prove that there are only finitely many sets of finite rank" ... but don't all natural numbers in the standard construction have finite rank? There are definitely infinitely many natural numbers (although each of them is of course a finite set in the standard construction). –  celtschk Apr 21 '14 at 20:20
OH. I meant to write "of a given finite rank". –  Asaf Karagila Apr 21 '14 at 20:27
Ah, OK, that makes more sense. Now let me get back to understanding the rest. :-) –  celtschk Apr 21 '14 at 20:28
OK, I think I see now why your proof works, but I don't see where you use induction. As far as I can see you just build a contradiction by proving $\rank(\bigcup x)<\rank x$, which contradicts $\bigcup x=x$. –  celtschk Apr 21 '14 at 20:35
Well, the $\rank$ function is a function. So if $\bigcup x=x$, how can it have two different ranks? –  Asaf Karagila Apr 21 '14 at 20:37

Let $A$ be set with $A\ne\emptyset$ and $A=\bigcup A$. We can define the notion of ordinal, i.e. transitive well-ordered (by $\in$) sets. Even without the Axiom of Infinity, the class of ordinals is proper and also well-ordered, hence there exists a smallest ordinal $\alpha$ that does not allow an injective map $\alpha\to A$. Then $\alpha$ cannot be finite, hence infinite ordinals exist. Hence there exists a smallest infinite ordinal $\omega$.

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