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:
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Picking a rational at each discontinuity
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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
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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!
Best Answer
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 $(\sup_{x<c}f(x),\inf_{x>c}f(x))$.