Jtdylktuy

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 $lim_{xto x_0}f(x)neq f(x_0)$. Denote
$$f(x_0^-)=lim_{xto x_0^-}f(x)$$
$$f(x_0^+)=lim_{xto x_0 ^+}f(x)$$

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)
$$S(x_i)={ymid f(x_i^-)<y<f(x_i^+)}$$
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 $qinmathbb{Q}$ thus making an injection $psi:bigcup S(x_i)rightarrow mathbb{Q}$. Thus $bigcup S(x_i)$ is at most countable.

My questions:

1. 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?

$f$ 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.