Proving without the Axiom of Choice (A.C.) that increasing real functions have countable discontinuities

Question

I know three different but similar proofs of the statement:

If $f:\mathbb{R}\to\mathbb{R}$ is an increasing function, then there are at most countably many discontinuities.

But each of the proofs relies on A.C. Therefore I am wondering if there is a way to prove this without using A.C.

The three proofs I know are as follows:

1. Picking a rational at each discontinuity

2. Proving that there are at most countably many discontinuities in each interval $[n,n+1]$, and then showing that this countable union of countable sets is countable

3. Using that there are at most countably many disjoint open intervals in $\mathbb{R}$, and that $(f(x-),f(x+)),x\in\{f\text{ is discontinuous at }x\}$ is a collection of disjoint intervals.

Please advise me about this!

Answer 1

You don't need AC for the proof n°1 (or n°3, I'm not exactly sure what's distinguishing them) because the rationals are well-orderable (this can be done without AC), therefore you can consider a well-ordering $(\Bbb Q,\preceq)$ and assign to each jump discontinuity the $\preceq$-least rational in $(\sup_{x<c}f(x),\inf_{x>c}f(x))$.

Answer 2

Partition $\mathbb{R}$ into disjoint bounded intervals $(I_m)_{m\geqslant1}.$ For example, $[0, 1), [-1, 0), [1, 2), \ldots$; the details don't matter. Take any strictly decreasing sequence $(c_n)_{n\geqslant1}$ with limit $0$ (again, the details don't matter), and partition $\mathbb{R}_{>0}$ into the disjoint intervals $J_1 = [c_1, +\infty),$ $J_n = [c_n, c_{n-1})$ for $n \geqslant 2.$ Then for all $m \geqslant 1$ and all $n \geqslant 1,$ the set $\{x \in \mathbb{R} : x \in I_m, \ f(x+) - f(x-) \in J_n\}$ is finite. Every point of discontinuity belongs to one of these finite sets, which can be arranged into a sequence, for example by a diagonal traversal.