TITLE: Constructing element which does not belong to a set
QUESTION [0 upvotes]: 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.

REPLY [2 votes]: 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 $A$).

REPLY [2 votes]: 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 $x$.
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