# Proving the Axiom of Choice for countable sets

I am new to the axiom of choice, and currently working my way through some exercises. I am struggling with the following exercise:

Exercise - Prove the Axiom of Choice (every surjective $f: X \to Y$ has a section) in the following two special cases:

1. Y is finite
2. X is countable

(A section has been previously defined as a function $s: Y \to X$ such that $f(s(y)) = y$ for all $y \in Y$)

My confusion - Now from what I understand, in (1) you use surjectivity of $f$ to pick an $x \in X$ such that $f(x) = y_0$ proving the case $|Y| = 1$, and then use induction to proof it for general $|Y| = N$.

I am a bit confused at (2) though. Would it be legal to take the previous argument and take the limit $N \to \infty$? I tried to google it but I got more confused after reading this question and seeing other references to the axiom of countable choice, suggesting that this result cannot be proven.

By the way, I am doing 'naive' set theory here; ZF/ZFC axiom systems and those kind of things have not been discussed in the course.

• Your solution for $(1)$ works. You need choice in general because you need a choice function on $Y$ to pick $s(x)$ among $f^{-1}(\{x\})$ for each $x \in X$. If you have a choice function $C$ then $s(x) = C(f^{-1}(\{x\}))$ works, right? Now how can you find such a choice function when $Y$ is countable? – nombre Dec 2 '15 at 15:51
• Ah can you just ennumerate $X$ as $x_0, x_1, ...$ and then take the lowest $n \in \mathbb{N}$ such that $x_n \in f^{-1}(\{x\})$? – Scipio Dec 2 '15 at 15:56
• But then what is all this about the Axiom of Countable Choice? – Scipio Dec 2 '15 at 15:57
• See Asaf Karagila's answer below. – nombre Dec 2 '15 at 16:18
• "Would it be legal to take the ..." No idea what you mean by legal, don't use that phrasing, it is meaningless. Perhaps you mean "valid". In that case, no, of course it is not valid. There is no sensible limit here. – Andrés E. Caicedo Dec 2 '15 at 16:34

The countable axiom of choice refers to the case where $Y$ is countable, so it is kind of different.

Let us see how we could do this. Since $X$ is countable there is a bijection $\phi$ with $\mathbb{N}$. Now you have a well defined order on $X$ :

$$x\leq y \text{ by definition if } \phi(x)\leq \phi(y)\text{ in }\mathbb{N}$$

I claim that this order on $Y$ verifies that the property " any non-empty set has a least element" is true (I don't remember right now the name of this property...).

Another way to state this is that in any non-empty $C$ set of $Y$ you can canonically $\textit{ choose }$ an element in $C$ namely $min(C)$. Apply this to construct the section for $f$.

• This is called the well-ordering property. – vadim123 Dec 2 '15 at 16:05

There is a terminological discrepancy here.

The Axiom of Countable Choice refers to the statement "If $S$ is a countable family of non-empty sets, then $S$ admits a choice function".

What you are trying to prove is that if $f\colon X\to Y$ and $X$ is countable, then $f$ admits a section.

This is not the same as choosing from a countable family of non-empty sets. If $\{A_n\mid n\in\Bbb N\}$ is a countable family of non-empty sets then a choice function would be a function from $\Bbb N$ into $\bigcup A_n$, rather than a function from $\Bbb N$ onto the union. The section map, if anything, would map $a\in\bigcup A_n$ to the least $n\in\Bbb N$, such that $a\in A_n$.

But an even more general thing is true. If $X$ can be well-ordered (so it can be uncountable as well), then by fixing a well-ordering of $X$ we can choose for each $y\in Y$ the minimal element of $\{x\in X\mid f(x)=y\}$ (which is non-empty by the surjectivity of $f$).