# Prove there are countable many discontinuities

Let I be a non-empty Interval and $f:I\rightarrow \mathbb{R}$ a monoton function. Show that $f$ has countable many Classification of discontinuities.

My prove is nearly complete. But I'm missing one step. I showed that there is a injection between the set of points where $f$ is not continuous and the rationals. According to my assistant professor I'm only allowed to do that, if I show that there are rationals. I know I can do that by applying the Archimedean property but for that I need to show ( formally correct ) that for $f_−(x) = sup f(x′) | x′ ∈ [a,b], x′ < x$ ,

$f_+(x) = inf (f(x′)) | x′ ∈ [a,b], x′ > x$,

where x is my point of Classification of discontinuities, that $f_+(x)<f_−(x)$

For me it's clear that $f_-(x)\leq f_+(x)$ can't be true, otherwise it wouldn't be a point of discontinuous so it has to be $f_-(x) < f_+(x)$ but how can this be proved?

• Don't you mean $f_+(x) > f_-(x)$? – Bungo Oct 31 '17 at 20:02
Let $D$ be the set of point in which $f$ is not continuous and let $x \in D$. Without loss of generality, assume that $f$ is increasing and let $f(x^+)$ and $f(x^-)$ be the right and left hand-side limits of $f$ respectively. Since $\mathbb{Q}$ is dense in $\mathbb{R}$ (and therefore in $I$), there exists $y \in \mathbb{Q}$ such that $f(x^-)<y<g(x^+)$. Since this can be done for every $x \in D\subset I$, one can set $g(x)=y$. For $x_1<x_2$, one has that $f(x_1^+) \leq f(x_2^-)$, hence $g(x_1) \neq g(x_2)$ for $x_1 \neq x_2$. This shows that $g:D \to \mathbb{Q}$ is an injection. Since $\mathbb{Q}$ is countable, then $A$ is countable
• Since $f$ is monotone, it also can have what is called jump discontinuities, ($f$ is well-defined in $I$). Since the limit at $x$ of $f$ does not exists, for every $x \in D$, it follows that $f(x^-)<f(x^+)$. – Adrián Naranjo Oct 31 '17 at 20:43
• assistant professor says no. We didn't define classification of discontinuities ,jump discontinuities nor limits. He said I still need to prove that $f_-(x)$ is strict smaller than $f_+(x)$ and I can do that with the property of the supremum and infimum. It's probably so simple and evident that I don't see it. Otherwise the proof is correct – the_asdf_word Nov 1 '17 at 19:59