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
$$f(x-)<r(x)<f(x+)$$
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?
Thanks in advance
Best Answer
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.