Assume that $f$ is an increasing function defined on $(a,b)$.
(i) Assume that $a \lt c \lt d \lt b$. Prove that $f(c^+) \le f(d^-)$.
My idea :
$f(c^+)=\inf \{f(x):x \in (a,b) \space \land x \gt c \}$
$f(d^-)=\sup\{f(x):x \in (a,b) \land x \lt d \}$
So, the more we get close to $c$ from right, the more $f$ decreases ... the more we get close to $d$ from left, the more $f$ increases ... so because $c \lt d \implies f(c) \le f(d)$ , the statement is proved. ( Is it true ? )
(ii) Assume that $D$ is the set of all points in $(a,b)$ such that $f$ is discontinuous on them. For each $c \in D$, choose a rational number $r_c$ from the interval $(f(c^-),f(c^+)$). Prove that if $c,d$ are two distinct points in $D$, $r_c \neq r_d$.
I have no idea about this part ...
(iii) With the use of part (ii), find a bijection from $D$ to a subset of rational numbers. Conclude that $D$ is countable.
Since i can't prove the previous part, so i can't define that function either ...
Notice that i wan't to use this method. I know there are many other methods to prove the main theorem. But i don't want them.
Any help would be great !
(i)
Consider $x\in (c,d).$ We have that $f(c^+)\le f(x) \le f(d^-).$
(ii)
We have that $$r_c<f(c^+)<f(d^-)<r_d$$ from where $r_c\ne r_d.$
(iii)
We define $c\in D\to r_c\in \mathbb{Q}\cap (f(c^-),f(c^+)).$ It is injective and thus we conclude that $D$ is countable.