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

  • $\begingroup$ Irrationals aren't countable, so such choice does not help you in proving the theorem $\endgroup$ – user160738 Apr 16 at 6:56
  • $\begingroup$ @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? $\endgroup$ – RScrlli Apr 16 at 6:58
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    $\begingroup$ 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 $\endgroup$ – user160738 Apr 16 at 7:04
  • $\begingroup$ @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. $\endgroup$ – RScrlli Apr 16 at 7:06
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    $\begingroup$ 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 $\Bbb Q\cap (f(x^-),f(x^+)).$ $\endgroup$ – 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.


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