Prove there are countable many discontinuities

Question

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

My prove is nearly complete. But I'm missing one step. I thought I showed that there is a injection between the set of points where $f$ is not continuous and the rationals, but according to my assistant professor, I'm only allowed to do that if I show that there are rationals in the interval $(f_-(x),f_+(x))$, for $$ f_−(x) = \sup \{f(x′) | x′ ∈ [a,b], x′ < x\}, $$

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

I know I can do that by applying the Archimedean property, but for that I need to show ( formally correct ) that $f_-(x)<f_+(x)$, where x is my point of discontinuity.

For me it's clear that $f_-(x)= f_+(x)$ 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?

thanks in advance

Answer 1

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

Answer 2

Since $f$ is monotone (non-decreasing I guess), it is clear that $f_-(x)\leq f_+(x)$ for every $x$. In fact, by definition of sup/inf, for an arbitrary $\epsilon>0$ you can find points $x_1<x<x_2$ such that $$ f_-(x)\geq f(x_1)>f_-(x)-\epsilon, $$ $$ f_+(x)\leq f(x_2)<f_+(x)+\epsilon. $$ Now, if by contradiction $f_-(x)>f_+(x)$, if you choose $0<\epsilon<(f_-(x)-f_+(x))/2$, you get $$ f(x_1)>f_-(x)-\epsilon > (f_-(x)+f_+(x))/2, $$ $$ f(x_2)<f_+(x)+\epsilon < (f_-(x)+f_+(x))/2. $$ The above inequalities imply $f(x_1)>f(x_2)$, which is not possible.

We now want to claim that if $x$ is a point of discontinuity, then it must be $f_-(x)<f_+(x)$. Equivalently, we show that if $f_-(x)=f_+(x)$, then $f$ is continuous at the point $x$.

So, assume $f_+(x)=f_-(x)$ and define $L:=f_+(x)=f_-(x)$. Let $\epsilon>0$. As before, there are $x_1<x<x_2$ such that $$ L\geq f(x_1)>L-\epsilon, $$ $$ L\leq f(x_2)<L+\epsilon $$ (I simply substituted $f_+(x),f_-(x)$ with $L$). Let $y\in[x_1,x_2]$. By monotonicity, $L-\epsilon <f(x_1)\leq f(y)\leq f(x_2)<L+\epsilon$, i.e., $f(y)\in (L-\epsilon,L+\epsilon)$ for every $y\in [x_1,x_2]$. So we see that the definition of continuity of the function $f$ at the point $x$ is satisfied, by defining $\delta(\epsilon)>0$ so that $(x-\delta(\epsilon),x+\delta(\epsilon))\subset[x_1,x_2]$, e.g. by choosing $\delta(\epsilon)=\min(|x-x_1|,|x-x_2|)$.

It's a bit annoying to prove these things by hand without the definition of limits and the classification of discontinuities, but it is definitely something one has to learn in math.