Theorem: Let $f$ be a real valued function of real variable defined on open interval $(a,b)$ and let $f$ be monotonic. Then the set of all discontinuities is at most countable.
I would like an explanation for several steps in the proof given here
The proof:
WLOG let $f$ be monotonic increasing and let $x_0$ be a point of discontinuity, because $f$ is discontinuous at $x_0$, it follows that $\lim_{x\to x_0}f(x)\neq f(x_0)$. Denote
$$f(x_0^-)=\lim_{x\to x_0^-}f(x)$$
$$f(x_0^+)=\lim_{x\to x_0 ^+}f(x)$$
Because $f$ is increasing, it follows that $f(x_0^-)<f(x_0^+)$ (Because $f$ is discontinuous, the strict ienquality holds.) Define intervals for each $x_i$ (the point of discontinuity)
$$S(x_i)=\{y\mid f(x_i^-)<y<f(x_i^+)\}$$
Now, the system of $S(x_i)$'s is pairwise disjoint and each of $S(x_i)$ is contained in compact interval $[f(a),f(b)]$. Now, it sufficies to pick a random rational number from each $S(x_i)$ thus associating each $x_i$ with some number $q\in\mathbb{Q}$ thus making an injection $\psi:\bigcup S(x_i)\rightarrow \mathbb{Q}$. Thus $\bigcup S(x_i)$ is at most countable.
My questions:
- What about the case when the one sided limits would go to infinity?
Is this a rigorous-enough explanation?
It is clear that if any left-sided limit goes to $+\infty$ at some point, say $x_i$, then in the interval $(x_i,b)$ is atleast one more point, say $r_1$ for which the value $f(r_1)>\infty$ (because $f$ is monotonic increasing), but that's not possible, since nothing is above $\infty$.
Simillarly, if any left-sided limit goes to $-\infty$ at some point, say $x_j$, then in the inrval $(a,x_i)$ there is some $r_2$ such that $f(r_2)<-\infty$…etc.
Is my reasoning correct?
Best Answer
$f$ is given to be real valued. There is no way $f(x-)$ or $f(x+)$ can be $\infty$ or $-\infty$ at any point. For example $f(y)\leq f(x)$ if $a<y<x$ so $f(x-) \leq f(x)<\infty$ . Similarly, $f(x-) \geq f(y)>-\infty$ for any $y \in (a,x)$, etc. Your argument for countability of the set of discontinuity points is correct.