$ \newcommand{\N}{\mathbb{N}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\R}{\mathbb{R}} $ I thought I understood AC (axiom of choice), but I am now totally confused through some discussions.

Let me examine whether each of the following examples needs AC. Assume ZF.

  1. Choose $ x \in (0, 1) $.
    AC is unnecessary. (One existential quantification over an interval.)

  2. Let $ n \in \N $. Choose $ x \in (0,1)^n $.
    AC is unnecessary because $x$ can be chosen by finite existential quantification.

  3. Let $ A = \{[a, b]| a,b \in \R,~a\le b\} $. There is a function $f:A \to \R$ such that $ f(I) \in I$ for all $ I \in A$.
    AC is unnecessary because we can declare a choice function $c: A \to \R$ defined by $ c(I) = \min I \in I $ for all $I \in A$ though $A$ is infinite.

  4. Let $ A = \{(a, b)| a,b \in \R,~a < b\} $. There is a function $f:A \to \R$ such that $ f(I) \in I$ for all $ I \in A$.
    AC is necessary because there is no finite quantification over $ A $ that specifies the choice of $x$.

  5. Let $ A = \{[a, b] \cap \Q| a,b \in \R,~a < b\} $ and $ B = \{(a, b) \cap \Q| a,b \in \R,~a < b\} $. Either there is a function $f:A \to \R$ such that $ f(I) \in I$ for all $ I \in A$ or there is a function $f:B \to \R$ ....
    AC is necessary by the same reasoning in 4.

  6. Let $ A = \{ (a,b)|a,b \in \Q,~a < b \} ​$. There is a function $f:A \to \R$ such that $ f(I) \in I$ for all $ I \in A$.
    AC is necessary by the same reasoning in 4 though $A$ is countable. I considered this example from the discussion by Karagila, which seems to be inconsistent with what I learned. Does countability assume some important role here?

Would you figure out what I am missing about the notion of AC?


I think I overlooked some simple factors in making examples. :) I would like to add one more example:

  1. Let $A \subseteq P(\R)$. There is a function $f:A \to \R$ such that $ f(I) \in I$ for all $ I \in A$.

Then, is 7. the only case AC actually needed?

  • $\begingroup$ How big the set is (provided it is not empty), or the nature of its elements, is irrelevant for choosing from finitely many sets. Countability of the sets you are choosing from plays no role. $\endgroup$ Nov 29 '21 at 4:33
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    $\begingroup$ If what you are trying to do in 4 is to define a choice function (as opposed to an element) for the family $A$, then it is still false that you need AC. You can define a choice function by letting $f((a,b)) = (a+b)/2$ (once you fix it so that the intervals are never empty). If that is what you are trying to do, then 6 also does not require AC, once fixed to $a\lt b$, with the same choice function. 5 may need it, though. $\endgroup$ Nov 29 '21 at 4:36
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    $\begingroup$ It takes quite some experience and ingenuity to come up with any example of a single concretely defined set that you definitely need AC for. Your (7) with $A=\mathcal P(\mathbb R)$ is probably the ur-example but can feel a bit unsatisfactory since it's rare in applications that you actually need the entire $\mathcal P(\mathbb R)$; another one that's a bit more motivated would be to let $A$ be the set of equivalence classes in $\mathbb R$ under the relation "$a\sim b$ iff $a-b\in \mathbb Q$. $\endgroup$ Nov 29 '21 at 4:54
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    $\begingroup$ You need to be careful about emptiness: in $7$ we need $A\subseteq\mathcal{P}_{\color{red}{\not=\emptyset}}(\mathbb{R})$. If $\emptyset \in A$ then $A$ has no choice function, but that has nothing to do with $\mathsf{AC}$. $\endgroup$ Nov 29 '21 at 5:02
  • $\begingroup$ Perhaps it would be more robust against the kind of errors @Noah points out if one defined a "choice function" to be one for which $f(x)\in x \lor x=\varnothing$ for every $x\in\operatorname{Dom}f$. $\endgroup$ Nov 29 '21 at 5:22

In fact $\mathsf{AC}$ is unnecessary in each of the examples in your post (before the edit that added example 7). I'll explain $5B$; in fact, the same choice function works for the other examples you ask about.

(I'm assuming that the "$\le$"s should be "$<$"s when open intervals are concerned, since $(a,a)=\emptyset$ and that has nothing to do with choice.)

The key point isn't that $\mathbb{Q}$ is countable per se, but rather that it is well-orderable (which is weaker). Specifically, fix at the outset a bijection $f:\mathbb{N}\rightarrow\mathbb{Q}$. Now given $a<b$ let $$c_{a,b}=f(\min\{n: f(n)\in (a,b)\}).$$

It's easy to see that $c_{a,b}\in(a,b)\cap\mathbb{Q}$ whenever $a<b$. Consequently the map sending the interval $(a,b)$ to the number $c_{a,b}$ is - modulo appropriate choice of $f$ - a "simply-definable" choice function for the family of intervals in $5B$, and so choice plays no role.

  • $\begingroup$ I guess this is the answer I was looking for! But does your statement assume the well-ordering theorem? (of course, it would be awkward if it does.) $\endgroup$
    – Hermis14
    Nov 29 '21 at 5:02
  • $\begingroup$ @Hermis14 No, why would it? Is there a point where you think it does? (Note that getting a bijection $\mathbb{N}\rightarrow\mathbb{Q}$ doesn't require choice at all, although any such bijection will be a bit messy.) $\endgroup$ Nov 29 '21 at 5:02
  • $\begingroup$ OK. I got it. Is $f$ the so-called enumeration of $Q$? $\endgroup$
    – Hermis14
    Nov 29 '21 at 5:04
  • $\begingroup$ @Hermis14 $f$ is an enumeration of $\mathbb{Q}$, it's not unique. (Actually all we need is that $f$ be a surjection, but it's a bit more natural to ask for a bijection). $\endgroup$ Nov 29 '21 at 5:07

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