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The standard axiom of choice can be expressed in the first order language of ZF by $$\forall X\big((X\neq\emptyset\wedge\emptyset\notin X)\implies\exists f:X\to\bigcup X\ \forall Y\in X(f(Y)\in Y)\big).$$ Naively, a simpler version would seem to be $$\forall X(X\neq\emptyset\implies\exists f:\{X\}\to X)$$ since we then have that $\forall A\in\{X\}(f(A)\in A)$, as specified explicitly in the choice axiom. This version is provable in ZF according to the discussion on this answer, however, and is thusly not an axiom of choice. How exactly do we lose choice moving from the first formulation to the second?

Alec Rhea
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    Your version just says that given any non-empty set, we can choose an element of that set. That’s true anyway, without the axiom of choice. – Brian M. Scott Mar 04 '21 at 01:32
  • But if we can choose an element of any non-empty set $Y$, call it $a_Y$, can't we construct a choice function $c:X\to\bigcup X$ for any family of nonempty sets $X$ by $\forall X\in Y(c(X)=a_X)$ and thusly get the standard axiom of choice? – Alec Rhea Mar 04 '21 at 01:36
  • @Alec: That question has been thoroughly asked before on the site. – Asaf Karagila Mar 04 '21 at 01:44
  • @AsafKaragila The functions we have aren't uniform if the quantifiers are in the order I have them above, I see that now thanks to your answer below. I was definitely slow on the uptake here haha. – Alec Rhea Mar 04 '21 at 01:46
  • To name a few, https://math.stackexchange.com/a/1839929/622 https://math.stackexchange.com/q/2515868/622 https://math.stackexchange.com/q/1877364/622 – Asaf Karagila Mar 04 '21 at 01:46
  • @AsafKaragila Wait... now I'm curious again, after reading that first link. If we extend the first-order language of ZF to contain infinitary sentences, do we get choice since we can now exhaustively existentially quantify for infinite sets? I think I remember something about second order set theory and set theory with infinitary sentences being equivalent in some sense, but would we still have to switch the order of the quantifiers? – Alec Rhea Mar 04 '21 at 01:50
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    Alec, there is more to just adding longer conjunctions, you need to worry about quantifiers, about deduction rules, etc. In principle, you'd need a proper class theory, quantifying over all sets that choice functions exist. Just a single classical first order axiom would be easier. – Asaf Karagila Mar 04 '21 at 02:02

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This is not the axiom of choice, since you're only requiring that given a single set, there is a function that selects an element from that set. It's just how we interpret quantifiers: first we replace $X$ with an instance, then we replace $f$ with an instance; but there's no requirement that these replacements are coherent.

Therefore, this is quite literally existential instantiation. This follows from the rules of logic, so it is provable in any weak theory you'd like (as long as you can discuss choice function in that theory).

If you switch the order of the quantifiers, $\exists f\forall X$, then you'd get something more akin to global choice, presuming you're working in second-order set theory.

Asaf Karagila
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  • Thank you; the definition with quantifiers switched wouldn't be valid in first order ZFC? Is it because $X$ is a second order object? Scratch that, $X$ is just a set, so is $f$ the second order thing we're quantifying over? I can't see where second order set theory comes into play, apologies if I'm being dense. Wait, is it because the codomain of $f$ has an element in every set of the universe and is thusly a subset of the universe, not an element of it? – Alec Rhea Mar 04 '21 at 01:38
  • Yes, $f$ would be second-order, since its domain would be the class of all sets (assuming you interpret this liberally when you write $f\colon{X}\to X$, otherwise no such $f$ can exist, since its domain would have to be different, and more importantly, pairwise disjoint sets). – Asaf Karagila Mar 04 '21 at 01:44