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:
- Y is finite
- 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.