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I've seen the axiom of choice as $\forall X\left[\varnothing \notin X\implies \exists f\colon X\rightarrow \bigcup_{A\in X}A,~~~ \forall A\in X\,(f(A)\in A)\right]$.

But isn't there a version which doesn't need the definition of a function and of the union?

Nathan Kaufmann
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  • Functions and unions (and $\emptyset$ for that matter) are just abbreviations for longer expressions in the language of set theory, so you can always just "unpack" them - it's just that the expression you get will be longer and harder to read. – Noah Schweber Feb 28 '24 at 17:22
  • Yes and this is what I'm looking for, where to find an unpacked version, or how to form one by myself. – Nathan Kaufmann Feb 28 '24 at 17:24
  • Hi! You can read Math.SE questions and answers on the empty set, the axiom of union, and functions, to see how to do each thing you want. The questions are not only about those things, but you should be able to glean what you need. As @NoahSchweber points out, this will be difficult to read, but if you don't care, have at it! – Brian Tung Feb 28 '24 at 17:31
  • Are you happy that you could write down a formula to express "$f$ is a function from $X$ to $Y$" or "$Y$ is equal to $\bigcup_{A \in X} A$" or "the value of $f(A)$ is $B$" and that sort of thing? The other "trick" you need is to sometimes add a new quantifier - for example, you can "unpack" a formula like $\varphi(f(A))$ to $(\exists B)((\text{the value of $f(A)$ is $B$}) \land \varphi(B))$. I think this answer might also be worth having a look at. – Izaak van Dongen Feb 28 '24 at 17:34
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    I think Zermelo used a variant with choice set. That's a formulation where you have a set $S$ of non-empty mutually disjoint sets, and you want a set $C$ that intersects each member of $S$ in exactly one member. It doesn't use function or union. The functional formulation can be derived from it using Cartesian products and union. – Chad K Feb 28 '24 at 18:39
  • $\in$ is not an atomic formula. $x\in x$, $x\in y$, $u\in x$ etc. are atomic formulas. – user14111 Feb 28 '24 at 18:55

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If $a$ is a set of pairwise disjoint sets, then there is a set $c$ whose intersection with each nonempty element of $a$ is a singleton. $$\forall a\{\exists s\exists t\exists x\exists y[s\in a\land t\in a\land x\in s\land x\in t\land y\in s\land y\notin t]\lor\exists c\forall s\forall w[s\in a\land w\in s\to\exists x\forall y(y\in c\land y\in s\leftrightarrow\forall z(z\in x\leftrightarrow z\in y))]\}$$

user14111
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