Szekeres problem 1.14

Question

I'm having trouble understanding Peter Szekeres's solution for problem 1.14 of his text "A Course in Modern Mathematical Physics". The question asks "If $f : [0, 1] \to \mathbb{R}$ is a non-decreasing function such that $f(0) = 0$, $f(1) = 1$, show that the places at which $f$ is not continuous form a countable subset of $[0, 1]$."

The official solution is as follows (my difficulty is with the very last part):

Let $f_{-}$ and $f_{+}$ be the functions defined by

$$ \begin{equation*} f_{-}(x) = \underset{y < x}{\mathrm{sup}}\, f(y) \qquad \text{and} \qquad f_{+}(x) = \underset{y > x}{\mathrm{inf}}\, f(y). \end{equation*} $$

Since the function is non-decreasing it is clear that $f_{-}(x) \leq f(x)$ and $f_{+}(x) \geq f(x)$ for all $x \in [0,1]$. The function $f$ is continuous at $x$ iff the limit both from the left and right is $f(y) \to f(x)$ as $y \to x$, from which it follows that $f$ is continuous at $x$ if and only if $f_{-}(x) = f(x) = f_{+}(x)$. If we set $\Delta f(x) = f_{+}(x) - f_{-}(x)$ then $\Delta f(x) \geq 0$ for all $x \in [0,1]$. At the points where $f$ is continuous $\Delta f(x) = 0$, while at points of discontinuity $\Delta f(x) > 0$.

For any finite sequence of points $x_1, x_2, \ldots, x_m$ it is evident that $\sum_{i=1}^m \Delta f(x_i) \leq 1$. Hence, there can be at most one point such that $\frac{1}{2} < \Delta f(x) \leq 1$, at most 2 points such that $\frac{1}{4} < \Delta f(x) \leq \frac{1}{2}$, and in general at most $2^n$ points such that $2^{-(n+1)} < \Delta f(x) \leq 2^{-n}$. Since the points at which $f(x)$ is discontinuous must be such that $\Delta f(x)$ falls in one of these half-open intervals of successively decreasing upper bound $2^{-n}$, we can arrange them in a sequence such that the $\Delta f(x)$ lie in intervals of successively decreasing upper bound $2^{-n}$, while arranging points of discontinuity for which $\Delta f$ lies in the same interval in, say, increasing order. All points of discontinuity of $f$ are thus set out in a sequence and must form a countable set.

Now I think this mostly makes sense. For example, in the statement

Since the points at which $f(x)$ is discontinuous must be such that $\Delta f(x)$ falls in one of these half-open intervals of successively decreasing upper bound $2^{-n}$, we can arrange them in a sequence such that the $\Delta f(x)$ lie in intervals of successively decreasing upper bound $2^{-n}$

$g = 2^{-n}$ defines a bijection between the set of natural numbers and the set of all interval upper bounds and so one discontinuity per interval is countable.

My difficulty is with the statement

while arranging points of discontinuity for which $\Delta f$ lies in the same interval in, say, increasing order.

I don't follow. Can't we also arrange the real numbers in increasing order? But of course the set of all real numbers is uncountable.

Answer 1

While we can arrange the real numbers in increasing order, we cannot list them in increasing order. If we start our listing by say $0, \frac{\pi-e}{100000},\dots$, and continue on listing some reals in increasing order, we're skill skip tons of numbers between $0$ and $\frac{\pi-e}{100000}$.

However, when there are only finitely many values to worry about, you can make a finite list consisting of those values. The 'increasing order' part is arbitrary, but just a way to specify a particular order of the list.

And since we can list the points of discontinuity, we know there are countably many.

(Another way to end the proof: The sets $S_n = \{x : \Delta f(x) \in (2^{-(n+1)}, 2^{-n}]\}$ are finite. The set $D$ of discontinuities is $\bigcup_{n=0}^\infty S_n$, the countable union of finite sets and thus finite.)