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In set theory, a set $A$ is a projective set if for any other sets $B, C$ and for any function $f:A\rightarrow B$ and surjective function $g:C\rightarrow B$, there exists a function $h:A\rightarrow C$ such that $g \circ h = f$. The Axiom of Choice is the statement that all sets are projective sets, while the weaker Axiom of Countable Choice is the statement that $\mathbb{N}$ is a projective set.

Suppose that the Axiom of Choice in classical set theory is replaced with the weaker axiom that the set of real numbers $\mathbb{R}$ is a projective set. Let us for the sake of this question call this new axiom the Axiom of Real Choice, for lack of a better term.

  1. Is the Axiom of Countable Choice provable from the Axiom of Real Choice? If so, then:
  2. Which results in mathematics (such as in real analysis, group theory, etc) that are proven using the full Axiom of Choice and cannot be proven using the Axiom of Countable Choice are now provable using the Axiom of Real Choice?
  3. Which results in mathematics that are proven using the full Axiom of Choice and cannot be proven using the Axiom of Countable Choice still cannot be proved using the Axiom of Real Choice?
  4. Are there any commonly used axioms that are inconsistent with the full Axiom of Choice that are consistent with the Axiom of Real Choice?
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    $\begingroup$ The answer to your first question appears to be yes. Assume the axiom of real choice. If $f:\Bbb N\to B$, define $\hat f:\Bbb R\to B:x\mapsto f(\lfloor x\rfloor)$. If $g:C\to B$ is surjective, there is an $\hat h:\Bbb R\to C$ such that $\hat f=g\circ\hat h$. Thus, $f(n)=\hat f(n)=g\big(\hat h(n)\big)$ for each $n\in\Bbb N$. Let $h=\hat h\upharpoonright\Bbb N$; then $f=g\circ h$. $\endgroup$
    – Brian M. Scott
    Commented Jul 27, 2020 at 19:58
  • $\begingroup$ @BrianM.Scott. We could instead take some (any) $b\in B$ (which exists, else $f $ doesn't exist) and let $f^*(x)=b$ for $x\in \Bbb R \setminus \Bbb N$ and $f^*|_{\Bbb N}=f.$ $\endgroup$
    – DanielWainfleet
    Commented Jul 27, 2020 at 21:32
  • $\begingroup$ @DanielWainfleet Are you proving, essentially, that every subset of a projective set is projective? $\endgroup$
    – Matemáticos Chibchas
    Commented Oct 6, 2020 at 19:06
  • $\begingroup$ @MatemáticosChibchas . I suppose so................. $\endgroup$
    – DanielWainfleet
    Commented Oct 7, 2020 at 16:52

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  1. Yes. If you have a countable family of sets, $\{X_n\mid n\in\Bbb N\}$, extend it to a family of size $\Bbb R$.

  2. Yes. "Every set of size $\Bbb R$ admits a choice function".

  3. The Axiom of Choice, and therefore all of its equivalents: e.g. every vector space has a basis; every commutative unital ring has a maximal ideal; etc. Including every equivalent of the Boolean Prime Ideal theorem (e.g. every commutative unital ring has a prime ideal).

  4. This is not particularly clear, but likely to be false.

This is Form 181 in the Howard–Rubin dictionary of choice principles. But the biggest problem with this principle is that it will follow from the conjunction of:

  1. $|\Bbb R|=\aleph_1$ and
  2. $\sf AC_{\aleph_1}$.

And while it's a somewhat nontrivial conjunction, it is also not particularly powerful or interesting (for example, it is easy to arrange a model where this conjunction holds, but there are sets which cannot be linearly ordered). You can find more information in the book, or in the online graphing website: https://cgraph.inters.co/.

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  • $\begingroup$ For 3., it would perhaps be more interesting to ask: Which theorems of analysis need more than real choice? $\endgroup$
    – Hagen von Eitzen
    Commented Jul 27, 2020 at 20:22
  • $\begingroup$ @HagenvonEitzen the reason why I phrased 2 and 3 the way I did is because there are mathematical theorems in other branches of mathematics that are unprovable with the axiom of countable choice that are countable with the full choice. $\endgroup$
    – user735141
    Commented Jul 27, 2020 at 20:42
  • $\begingroup$ @Hagen: Well, there are books upon books to fill with choiceless analysis. $\endgroup$
    – Asaf Karagila ♦
    Commented Jul 27, 2020 at 20:46
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    $\begingroup$ @MadeleineBirchfield: Yes, since this does not imply things like the Boolean Prime Ideal Theorem, you already miss out on a lot of them. $\endgroup$
    – Asaf Karagila ♦
    Commented Jul 27, 2020 at 20:47

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