I am currently reading Baby Rudin and I have some trouble understanding the proof
4.30 Theorem Let $f$ be monotonic on $(a, b)$. Then the set of points of $(a, b)$ at
which $f$ is discontinuous is at most countable.
Proof Suppose, for the sake of definiteness, that $f$ is increasing, and
let $E$ be the set of points at which $f$ is discontinuous.
With every point $x$ of $E$ we associate a rational number $r(x)$ such
that
$f(x-) < r(x) <f(x+)$.
Since $x_1 < x_2$ implies $f(x_1 +) \leq f(x_2 – )$, we see that $r(x_1) \ne r(x_2 )$ if
$x_1\ne x_2$ We have thus established a 1-1 correspondence between the set $E$ and
a subset of the set of rational numbers. The latter, as we know, is countable.
I don't understand why it is possible to associate a different $r(x)$ for all $x$ , I think the reason why a monotone function can have at most countable discontinuities is that for any uncountable set of positive real numbers $S \ \ \ \sum\limits_{s\in S}s$ diverge to infinity and since $\sum\limits_ {x \in E}f(x+) – f(x-) <f(b-)- f(a+)$ so it is not uncountable (this is my understanding of this theorem maybe it is wrong) and since this was not been in the proof means Rudin didn't use it so lets for a minute imagine that there is uncountable set of positive real numbers $S \ \ \ \sum\limits_{s\in S}s$ converge then we can make a monotone function with uncountably infinite discontinuities then it would be impossible to associate a different $r(x)$ for all $x$
My example about uncountable infinite discontinuities is not a proof but a part of the question, since Rudin didn’t address the fact that any sum of elements of uncountable infinite set of positive reals diverges means He didn’t use it in his proof and that means the proof works without it (which I don't think is right if it can make a monotone function with uncountable infinite discontinuities ). And here I am confused, since he didn’t address that and without mentioning this fact the proof is incomplete, except if the proof somehow implies that any uncountable infinite set of positive reals diverges.
Best Answer
The point is that if $f(x_+)>f(x_-)$ and $f$ is monotonic increasing, then there is no other point $x'$ such that $\big(f(x_-),f(x_+)\big)\cap \big( f(x'_-),f(x'_+) \big)\neq \emptyset$. Namely, any discontinuity has a corresponding nontrivial interval, corresponding to the jump in values, disjoint from the interval corresponding to any other discontinuity. If $r(x)$ would not be injective, then those intervals would have to intersect.
I think your alternative proof also works, but one must still argue why an uncountable sum of positive number must diverge.