1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.