# Discontinuity of monotonic function

I've seen this topic has been already discussed in this question but actually my doubt is slightly different so I consider opportune to ask it as a sigle question, please correct me if I am wrong.

Basically on Rudin's Principles of Mathematical Analysis (2º edition) there's a proof regarding the set of discontinuities $$(E)$$ of a monotonic function, that according to the theorem must be countable.

With every $$x\in E$$ we associate a rational number $$r(x)$$ such that $$f(x-)

Since f is monotonic both $$f(x−),f(x+)$$ exists and hence we can find such a rational number $$r(x)$$. Thus we have a $$1−1$$ correspondence between the set E and a subset of the set of rational numbers.

The idea of choosing a rational number seems quite arbitrary for me, and although I do see how this proves that the set $$E$$ is countable I don't see why we cannot choose, instead, a number $$i(x)\in \mathbb R \setminus \mathbb Q$$.

Since the irrational numbers are dense in $$\mathbb{R}$$ as well, we can find such a $$i(x)$$. (The important part is that the irrational numbers are actually uncountable) If I manage to obtain a $$1-1$$ correspondence between the discontinuity points and the irrational numbers, wouldn't that show that the set $$E$$ is actually uncountable?

Can someone explain me why this is a wrong approach?

• @user160738 indeed, but having a $1-1$ correspondence between the discontinuity points and the irrational numbers wouldn't show that the set is actually uncountable? – RScrlli Apr 16 at 6:58
• Indeed, it doesn't show that it's uncountable. But it doen't show it's countable as well, which was to be shown. I'm saying that there's no point in making such correspondence between $E$ and a subset of irrationals, which might or might not be countable – user160738 Apr 16 at 7:04
• @user160738 I was neglecting the fact that we are actually constructing a bijection between the set $E$ and a subset of $\mathbb{R}\setminus \mathbb{Q}$. Thanks for pointing that out. – RScrlli Apr 16 at 7:06
• This proof is shorter and simpler than others I've seen, and sweeter because it does not need a consequence of the Axiom of Choice (AC) that a countable union of countable sets is countable. Without AC we can define a well-order $<^*$ on $\Bbb Q$ and for $x\in E$ we can define $r(x)$ to be the $<^*$-least member of $\Bbb Q\cap (f(x^-),f(x^+)).$ – DanielWainfleet Apr 16 at 10:13
The method used by Rudin creates a bijection between $$A$$ and a subset of $$\mathbb Q$$. Since $$\mathbb Q$$ is countable, you deduce from it that $$E$$ is countable (or finite).
If you use $$\mathbb R\setminus\mathbb Q$$ instead of $$\mathbb Q$$, what you deduce is that $$E$$ is at most uncountable. That gives you no information.