0

If $A$ is an arbitrary set, is it possible to construct an element $a\not\in A$ without the use of the Axiom of Regularity?

  • It is clear that such $a$ exists (else $A$ would be the set of all sets), but this is not an explicit construction.

  • Under assumption of the Axiom of Regularity, one could choose $a=A$.

To give more context, I'm starting with a family $\{X_i\mid i\in I\}$ of non-empty sets, and would like to find a sequence $\langle y_i \rangle_{i\in I}$ such that $y_i\not\in X_i$ for each $i\in I$, without using the Axiom of Choice and the Axiom of Regularity.

Zuy
  • 4,962
  • 1
    What exactly does "construct" mean? It's not entirely clear to me what sentence exactly you're asking about here. For example, are you asking whether there is a single formula $\varphi(x,y)$ in the language of set theory such that $\mathsf{ZF-Reg}$ proves "For every $x$ there is exactly one $y$ satisfying $\varphi(x,y)$, and for this $y$ we have $y\not\in x$"? Or something else? – Noah Schweber Mar 07 '21 at 08:37
  • @NoahSchweber Yes, exactly! – Zuy Mar 07 '21 at 08:41
  • I hate this question. It's been asked to death, but it's hard to locate, so it's never obvious where the duplicates are. – Asaf Karagila Mar 07 '21 at 10:09
  • Your question is literally "is it possible in ZF-Reg that the set of all sets exists", and when you read it like that, it should be obvious how to answer it. – Asaf Karagila Mar 07 '21 at 10:10
  • Well, the answer seems obvious now, and I was kind of convinced that there should be a corresponding construction, but I was missing the right idea... Excuse me, sir. – Zuy Mar 07 '21 at 10:13
  • I've put a duplicate that is itself a duplicate, there are probably many others as well. – Asaf Karagila Mar 07 '21 at 10:16

2 Answers2

2

This might not be constructive enough, but there exists an element in the Hartogs ordinal of $A$ that is not a member of $A$. Otherwise it would be a subset of $A$, a contradiction (there can be no injection from the Hartogs ordinal into $A$).

Tesla Daybreak
  • 1,663
2

Here's one observation (which is essentially the same as Tesla Daybreak's answer):

$\mathsf{ZF-Reg}$ is strong enough to prove that no set contains every ordinal. Now given a set $x$, let $y$ be the least ordinal not in $x$; regardless of whether regularity holds in the universe, the ordinals themselves are well-ordered as usual. This is totally explicit and pins down $y$ uniquely in terms of $x$.

Noah Schweber
  • 260,658