# Prove there are countable many discontinuities

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), 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?

• Don't you mean $f_+(x) > f_-(x)$?
– user169852
Oct 31, 2017 at 20:02
• The proof is essentially four observations. (1) the only type of discontinuity that is possible for a monotone function is a jump discontinuity; (2) each jump corresponds to an interval in the codomain, consisting of the points that are "skipped"; (3) these intervals are pairwise disjoint; (4) each interval contains a rational.
– user169852
Oct 31, 2017 at 20:05
• yes you're right, it's my mistake Nov 1, 2017 at 20:08

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

• to use Rationals I need to apply the archimedean property. For that I need to show that f(x-) is strict smaller than f(x+) Oct 31, 2017 at 20:37
• 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 $f(x^-)<f(x^+)$. Oct 31, 2017 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 Nov 1, 2017 at 19:59

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 $$x_1 such that $$f_-(x)\geq f(x_1)>f_-(x)-\epsilon,$$ $$f_+(x)\leq f(x_2) Now, if by contradiction $$f_-(x)>f_+(x)$$, if you choose $$0<\epsilon<(f_-(x)-f_+(x))/2$$, you get $$f(x_1)>f_-(x)-\epsilon > (f_-(x)+f_+(x))/2,$$ $$f(x_2) The above inequalities imply $$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 $$f_-(x). Equivalently, we show that if $$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 $$x_1 such that $$L\geq f(x_1)>L-\epsilon,$$ $$L\leq f(x_2) (I simply substituted $$f_+(x),f_-(x)$$ with $$L$$). Let $$y\in[x_1,x_2]$$. By monotonicity, $$L-\epsilon , i.e., $$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.