# Prove that $f:(a, b) \to \mathbb R$ has at most countably many simple discontinuities

This is problem 17 in baby Rudin's chapter on continuity. He has a hint to use triplets of rationals that bound each simple discontinuity on the left, right, and in between the values of the limits from the left and right. It seems like this can be weakened to just rationals to the left and right.

Simple discontinuities are those in which the limit from the left and right exist, so there must be intervals to the left and right of a simple continuity on which no other simple discontinuity can exist. More precisely, let $c$ and $c'$ be simple discontinuities for $f$ on $(a,b)$ and consider the limit, $l$, of $f$ approaching $c$ from the left:

$$\forall \epsilon >0, \exists \delta>0 : c-x<\delta \Rightarrow |l-f(x)|<\epsilon$$

but $\exists \epsilon '>0 \forall \delta ' >0 : |c'-x|<\delta ' \Rightarrow |f(c')-f(x)|>\epsilon ' \\ \therefore \epsilon = \epsilon ' , \delta ' = \delta \rightarrow \leftarrow$

Therefore, you can make an injection from the set of simple discontinuities to a subset of rationals by associating each simple discontinuity with one rational in the aforementioned "free" interval to its left; from there compose with the map from rationals to integers to show countable. Is this argument correct?

so there must be intervals to the left and right of a simple continuity on which no other simple discontinuity can exist.

That is incorrect. Consider an enumeration $r_n$ of the rationals, and let

$$f(x) = \sum_{\substack{n \in \mathbb{N}\\r_n < x}} 2^{-n}.$$

Then $f$ is a strictly monotonic function that has a jump discontinuity in every rational number.

However, as the points of discontinuity approach any fixed $y \in \mathbb{R}$, the jumps tend to $0$, in fact, the sum of all jumps in the discontinuities $\neq y$ contained in a neighbourhood of $y$ tends to $0$ when the neighbourhood shrinks to a point.

Let $E_n = \left\{ x \in \left( a , b \right) \mid \lvert \lim_{t \to x^+} f(t) - \lim_{t \to x^-} f(t) \rvert > \frac{1}{n} \right\}$, the set of simple discontinuities whose jumps are greater than $\frac{1}{n}$. Then $E_1 \subset E_2 \subset\cdots$ and $\cup^\infty_{i=1} E_i$ is the set of all simple discontinuities. If all of the $E_n$ are countable, then $\cup^\infty_{i=1}E_i$ is countable.

Suppose $x \in E_n$.

The left-hand limit exists, so

$$\exists \delta^-_x > 0 \colon t \in \left( x - \delta^-_x, x \right) \implies \lvert f(t) - \lim_{t \to x^-} f(t) \rvert < \frac{1}{4n}$$

Suppose $y \in \left( x - \delta^-_x, x \right)$ is a simple discontinuity. Then

$$\exists \delta^-_y > 0 \colon t \in ( y - \delta^-_y, y ) \implies \lvert f(t) - \lim_{t \to y^-} f(t) \rvert < \frac{1}{4n} \\ \exists \delta^+_y > 0 \colon t \in ( y, y + \delta^+_y ) \implies \lvert f(t) - \lim_{t \to y^+} f(t) \rvert < \frac{1}{4n}$$

Then $\exists z^- \in (y - \delta^-_y, y) \cap (x - \delta^-_x, x), z^+ \in (y, y + \delta^+_y) \cap (x - \delta^-_x, x) \colon$

$$\lvert \lim_{t \to y^+} f(t) - \lim_{t \to y^-} f(t) \rvert \leq \\ \lvert \lim_{t \to y^+} f(t) - \lim_{t \to x^-} f(t) \rvert + \lvert \lim_{t \to y^-} f(t) - \lim_{t \to x^-} f(t) \rvert \leq \\ \lvert \lim_{t \to y^+} f(t) - f(z^+) \rvert + \lvert \lim_{t \to x^-} f(t) - f(z^+) \rvert + \lvert \lim_{t \to y^-} f(t) - f(z^-) \rvert + \lvert \lim_{t \to x^-} f(t) - f(z^-) \rvert \leq \\ \frac{1}{n}$$

So that

$$\forall x \in E_n \exists \delta > 0 \colon y \in (x - \delta, x ) \implies y \not \in E_n$$

This says that for a jump at $x$, there exists a left-hand neighborhood around x such that the jumps in the neighborhood are smaller than the jump at $x$ (we could prove the same thing for a right-hand neighborhood, but that turns out to be unnecessary). For each $x$, choose a rational number $q \in \mathbb{Q}$ from $(x - \delta, x )$, e.g.,

$$\frac{ \lceil (x - \delta) ( \lceil \frac{1}{\delta} \rceil + 1 ) \rceil + 1}{\lceil \frac{1}{\delta} \rceil + 1}$$

Since the $(x - \delta, x)$ are disjoint, $g \colon x \mapsto q$ is an injection $g \colon E_n \rightarrowtail \mathbb{Q}$ (each $x$ maps to a different $q$). Since $\mathbb{Q}$ is countable, so is $E_n$.

1. Jump discontinuities are at most countable

Let $J=\{x\in(a,b): f(x+0), f(x-0)\text{ both exist but are not equal }\}$, and $$J_+=\{x\in J:f(x-0)<f(x+0)\}, J_-=\{x\in J: f(x-0)>f(x+0)\}$$ So we have a decomposition $J=J_+\cup J_-$. We now prove that $J_+$ is at most countable.

For any $x\in J_+$, by the density of rationals, we can choose $r_x\in (f(x-0),f(x+0))\cap\mathbb{Q}$.

On the one hand, $f(x-0)<r_x$ we know $\exists \delta_-\in(0,x-a)$ such that $\forall z\in (x-\delta_-,x)$ we have $f(z)<r_x$. Also by the density of rationals, we can choose $s_x\in (x-\delta_-,x)\cap\mathbb{Q}$. As a result, $\forall z\in(s_x,x)$ we have $f(z)<r_x$.

On the other hand, $f(x+0)>r_x$ we know $\exists \delta_+\in(0,b-x)$ such that $\forall z\in (x, x+\delta_+)$ we have $f(z)>r_x$. Also by the density of rationals, we can choose $t_x\in (x, x+\delta_+)\cap\mathbb{Q}$.As a result $\forall z\in(x,t_x)$ we have $f(z)>r_x$.

Thus, $\forall x\in J_+$ there corresponds a triplet $(r_x,s_x,t_x)$. By Axiom of Choice, we can define a function: \begin{align*} j_+:J_+ &\to \mathbb{Q}^3\\ x &\mapsto (r_x,s_x,t_x) \end{align*} We show that it's an injection: Suppose $\exists x\neq y \in J_+$ such that $j_+(x)=j_+(y)$, i.e. $r_x=r_y=r,s_x=s_y=s,t_x=t_y=t$. Without loss of generality assuming $x<y$, note that we can find $z\in (a,b)$ satisfying $s<x<z<y<t$. From $x<z<t$ follows $f(z)>r$ and from $s<z<y$ follows $f(z)<r$, which is a contradiction.

Now we have an injection from $J_+$ to $\mathbb{Q}^3$ and by the countability of the latter, we conclude that $J_+$ is at most countable. The same argument goes for $J_-$ and proves $J_-$ and hence $J=J_+\cup J_-$ are at most countable.

2. Removable disconitnuities are at most countable

Let $R=\{x\in(a,b): f(x+0),f(x-0)\text{ exist and equal but differs from }f(x)\}$. Decompose this set into two: $R=R_+\cup R_-$, where $$R_+=\{x\in R: f(x+0)=f(x-0)>f(x)\},R_-=\{x\in R: f(x+0)=f(x-0)<f(x)\}$$ Now we show that $R_+$ is at most countable.

For any $x\in R_+$, let $A_x=\lim\limits_{t\to x}f(t)>f(x)$. By the density of rationals we can choose $r_x\in (f(x),A_x)\cap \mathbb{Q}$. We know $\exists \delta\in (0,\min\{x-a,b-x\})$ such that $\forall z\in (x-\delta,x+\delta)\backslash\{x\}$ we have $f(z)>r_x$. Also by the density of rationals we can choose $s_x\in (x-\delta,x)\cap \mathbb{Q}, t_x\in (x,x+\delta)\cap \mathbb{Q}$. As a result, $\forall z\in (s_x,t_x)\backslash \{x\}$ we have $f(z)>r_x$. By Axiom of Choice, we can define a function: \begin{align*} r_+:R_+&\to \mathbb{Q}^3\\ x&\mapsto (r_x,s_x,t_x) \end{align*}

Suppose that $r_+$ is not an injection. Then we can find $x<y\in R_+$ such that $r_+(x)=r_+(y)$, i.e. $r_x=r_y=r,s_x=s_y=s,t_x=t_y=t$. From $x\in (s,t)\backslash \{y\}$ follows $f(x)>r$ which contradicts $r=r_x\in(f(x),A_x)$. Now we have an injection from $R_+$ into $\mathbb{Q}^3$ and by the countability of the latter we know that $R_+$ is at most countable. By a similar argument, we can define an injection $r_-:R_-\to \mathbb{Q}^3$ and knows that $R_-$ is also at most countable. We conclude that $R=R_+\cup R_-$ is at most countable.

3. Simple dicontinuities are at most countable.

This follows from 1 and 2 since simple discontinuities are either jumping or removable.

• I really don't understand why someone downvoted my answer. :( – Mathis Dec 14 '17 at 1:25