$B\subset\mathbb{R}$ and $f:B\to\mathbb{R}$ is an increasing function. $f$ is continuous at every element of $B$ except for a countable subset of $B$.

Question

I am reading "Measure, Integration & Real Analysis" by Sheldon Axler.
The following exercise is Exercise 22 on p.39 in Exercises 2B in this book.

Exercise 22
Suppose $B\subset\mathbb{R}$ and $f:B\to\mathbb{R}$ is an increasing function. Prove that $f$ is continuous at every element of $B$ except for a countable subset of $B$.

I tried to solve Exercise 22 but I was not able to solve it.

I found the following:

lorenzo's question and his proof
Hagen von Eitzen's proof

I think lorenzo tried to fill the gap in Hagen von Eitzen's proof.

First I checked Hagen von Eitzen's proof:
Let $C:=\{x\in B:f\text{ is not continuous at }x\}.$
Let $b\in B.$
If $\{f(x):x\in B,x<b\}=\emptyset$ and $\{f(x):x\in B,b<x\}=\emptyset$, then $B=\{b\}$ and $f$ is continuous at $b$.
Let $b\in C.$
Then, since $f$ is not continuous at $b$, $\{f(x):x\in B,x<b\}\neq\emptyset$ or $\{f(x):x\in B,b<x\}\neq\emptyset.$
We prove the following inequality:
$$\sup\{f(x):x\in B,x<b\}<\inf\{f(x):x\in B,b<x\}.$$
For any $y\in\{f(x):x\in B,x<b\}$, $y\leq f(b)$ because $f$ is an increasing function.
So, $\sup\{f(x):x\in B,x<b\}\leq f(b).$
For any $y\in\{f(x):x\in B,b<x\}$, $f(b)\leq y$ because $f$ is an increasing function.
So, $f(b)\leq\inf\{f(x):x\in B,b<x\}.$
Therefore, $\sup\{f(x):x\in B,x<b\}\leq f(b)\leq\inf\{f(x):x\in B,b<x\}.$

(Case 1) We consider the case in which $\{f(x):x\in B,x<b\}=\emptyset$ and $\{f(x):x\in B,b<x\}\neq\emptyset.$
$\sup\{f(x):x\in B,x<b\}=-\infty$ and $f(b)<\inf\{f(x):x\in B,b<x\}$ (if $f(b)=\inf\{f(x):x\in B,b<x\}$, then $f$ is continuous at $b$. This is a contradiction.)
Therefore, in this case, $\sup\{f(x):x\in B,x<b\}< f(b)<\inf\{f(x):x\in B,b<x\}.$

(Case 2) We consider the case in which $\{f(x):x\in B,x<b\}\neq\emptyset$ and $\{f(x):x\in B,b<x\}=\emptyset.$
$\inf\{f(x):x\in B,b<x\}=\infty$ and $\sup\{f(x):x\in B,x<b\}<f(b)$ (if $\sup\{f(x):x\in B,x<b\}=f(b)$, then $f$ is continuous at $b$. This is a contradiction.)
Therefore, in this case, $\sup\{f(x):x\in B,x<b\}< f(b)<\inf\{f(x):x\in B,b<x\}.$

(Case 3) We consider the case in which $\{f(x):x\in B,x<b\}\neq\emptyset$ and $\{f(x):x\in B,b<x\}\neq\emptyset.$
$\sup\{f(x):x\in B,x<b\}<f(b)$ or $f(b)<\inf\{f(x):x\in B,b<x\}$ holds. (if $\sup\{f(x):x\in B,x<b\}=f(b)$ and $f(b)=\inf\{f(x):x\in B,b<x\}$, then $f$ is continuous at $b$. This is a contradiction.)
Therefore, in this case, $\sup\{f(x):x\in B,x<b\}\leq f(b)<\inf\{f(x):x\in B,b<x\}.$ or $\sup\{f(x):x\in B,x<b\}< f(b)\leq\inf\{f(x):x\in B,b<x\}.$

Therefore, $\sup\{f(x):x\in B,x<b\}<\inf\{f(x):x\in B,b<x\}$ holds.

Next I checked lorenzo's proof.

lorenzo wrote like the following:
$I_d=:(\sup_{x<d,\ x\in B}f(x),\inf_{x>d,\ x\in B} f(x))$ is a non-empty interval in $\mathbb{R}$ hence there must be a rational number $q_d$ in it which we can choose and since $I_d\cap I_{d'}=\emptyset$ for $d\neq d'$ $\cdots$

Let $B=(-\infty,-1]\cup [1,\infty)$.
Suppose $f$ is a real-valued function on $B$ such that $f(x)=x-1$ if $x<-1$ and $f(-1)=-1$ and $f(1)=1$ and $f(x)=x+1$ if $1<x.$
Let $d=-1$ and $d'=1$.
Then, $I_d=(\sup_{x<d,\ x\in B}f(x),\inf_{x>d,\ x\in B} f(x))=(-2,1).$
Then, $I_{d'}=(\sup_{x<d',\ x\in B}f(x),\inf_{x>d',\ x\in B} f(x))=(-1,2).$
So, $I_d\cap I_{d'}=(-1,1)\neq\emptyset.$
So, I think lorenzo's proof needs to be modified.

How to modify lorenzo's proof?

Answer 1

Here is a solution:

Since $f$ is an increasing function define on $B$, we have that, for all left limit point $d$ of $B$ there is $\lim_{x \to d^-}f(x)$ and, for all right limit point $e$ of $B$ there is $\lim_{x \to e^+}f(x)$.

Suppose $b \in B$ and $f$ is discontinuous at $b$ then either

1. $b$ is a left limit point of $B$ and $\lim_{x \to b^-}f(x)\neq f(b)$ or
2. $b$ is a right limit point of $B$ and $\lim_{x \to b^+}f(x)\neq f(b)$.

Of course, it is possible that 1 and 2 be both true. Note also that if $b$ is an isolated point (not left limit point nor right limit point) then $f$ is continuous at $b$.

For all $b \in B$ such that $f$ is discontinuous at $b$, let $I_b$ be the open interval defined as:

case 1. if $b$ is a left limit point of $B$ and $\lim_{x \to b^-}f(x)\neq f(b)$, then $\lim_{x \to b^-}f(x) < f(b)$ and we define $I_b=(\lim_{x \to b^-}f(x), f(b))$;

case 2. if $b$ is not a left limit point of $B$ or $\lim_{x \to b^-}f(x)= f(b)$, then, since $f$ is discontinuous at $b$, we must have that $b$ is a right limit point of $B$ and $\lim_{x \to b^+}f(x)\neq f(b)$. Then, $f(b)< \lim_{x \to b^+}f(x)$ and we define $I_b=(f(b), \lim_{x \to b^+}f(x))$

It is easy to prove that if $b, b' \in B$, $b<b'$ and $f$ is discontinuous at $b$ and $b'$, then $I_b \cap I_{b'}=\emptyset$. In fact, we have:

If both $b$ and $b'$ are in case 1, then $\lim_{x \to b^-}f(x)< f(b) \leqslant \lim_{x \to b'^-}f(x) <f(b')$, so $I_b \cap I_{b'}=\emptyset$.

If $b$ is in case 1 and $b'$ is in case 2, then $\lim_{x \to b^-}f(x)< f(b) \leqslant f(b')< \lim_{x \to b'^+}f(x) $, so $I_b \cap I_{b'}=\emptyset$.

If $b$ is in case 2 and $b'$ is in case 1, then $f(b)< \lim_{x \to b^+}f(x) \leqslant \lim_{x \to b'^-}f(x) <f(b')$, so $I_b \cap I_{b'}=\emptyset$.

If both $b$ and $b'$ are in case 2, then $f(b)< \lim_{x \to b^+}f(x) \leqslant f(b') < \lim_{x \to b'^+}f(x)$, so $I_b \cap I_{b'}=\emptyset$.

So we have proved that $I_b \cap I_{b'}=\emptyset$.

Now, let $C:=\{x\in B:f\text{ is not continuous at }x\}$. For each $b \in C$, choose $\varphi(b) \in I_b \cap \Bbb Q$. It follows immediately that $\varphi: C \rightarrow \Bbb Q$ is injective. So $C$ is countable.