real analysis - Countable subset and monotonic function

let E be subset of R which has no isloated points(or C does not have any isolated point of E) and C be countable subset of R does there exist a monotonic function on E which is continuous only at points in E-C?

The problem is from Royden 4th edition page 109.

I know the proof in case E is an open bounded interval only.

Answer

Let

$$L=\left\{x\in C:\exists y_x\in\Bbb R\big(y_x

this is the set of points in $C$ that are not limits from the left of points in $E$. Another way to say it is that $x\in L$ if and only if $x\in C$ and $x>\sup_{\Bbb R}\{y\in E:y

$$C\cup Y=\{x_n:n\in\Bbb N\}\;.$$

(I’m assuming that $C$ is countably infinite; if $C$ is finite, the problem is fairly trivial.) Let

$$f:E\cup Y\to\Bbb R:x\mapsto\sum_{x_n\le x}\frac1{2^n}\;.$$

Finally, let

$$g:E\to\Bbb R:x\mapsto\begin{cases}
f(y_{x_n}),&\text{if }x=x_n\in Y\\
f(x),&\text{otherwise}\;.
\end{cases}$$

The definition of $f$ ensures that $g$ is discontinuous from the left at every point of $C\setminus L$ and continuous everywhere else, and the modification to get $g$ ensures that $g$ is discontinuous from the right at every point of $L$ without affecting continuity at any other point of $E$. Thus, $g$ is discontinuous precisely at the points of $C$.