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!

  • $\begingroup$ How do you do 2. for an interval? I assume you mean an argument that somehow uses the boundedness of the interval and avoids choice. $\endgroup$ Feb 18, 2022 at 23:44
  • 1
    $\begingroup$ My idea was like this: if there were uncountable discontinuities in an interval $[n,n+1]$, then $f(n+1)-f(n)\geq \sum_{x\text{ is discontinuity point in }[n,n+1]} (f(x+)-f(x-))=\infty$ and reach a contradiction. $\endgroup$ Feb 19, 2022 at 0:12
  • $\begingroup$ When a proof tells you to "choose" something or to "pick" something, it may or may not mean "apply the axiom of choice". It may instead mean "It is left as an exercise to the reader to determine whether the choice may be made without the axiom of choice". $\endgroup$
    – Lee Mosher
    Feb 19, 2022 at 3:22

2 Answers 2


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 $(\sup_{x<c}f(x),\inf_{x>c}f(x))$.

  • 2
    $\begingroup$ You can be more explicit: pick the rational with smallest denominator, and among those, of smallest numerator, and among those, the smaller one. $\endgroup$ Feb 18, 2022 at 23:43

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.


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