Let $C$ be a countable subset of $(a,b)$. Then there is an increasing continuous function on $(a,b)$ that is continuous only on $(a,b)\setminus C$
This is an example from Royden's real analysis book. The function defined on $(a,b)$ is $f(x)=\sum_{\{n:q_n \le x\}}\frac{1}{2^n}$ where $\{q_n\}$ is an enumeration of $C$ . To show continuity on $(a,b)\setminus C$ he says let $x_0$ be in $(a,b)\setminus C$ and let $n$ be any natural number. Then there exists an interval, $I$, that contains $x_0$. In addition, $q_n$ is not in $I$ for $1\le k \le n$. Then this implys that $|f(x)-f(x_0)| < \frac{1}{2^n}$ for $x \in I$. This last sentence follows from the fact that if $a<u<v<b$ then $f(v)-f(u)=\sum_{\{n:u<q_n \le v\}}\frac{1}{2^n}$
Im having trouble understanding why such an $I$ exists. For example, $C$ was a countable and dense subset of $(a,b)$ then for any $n$ large enough so that $x_0 < q_n$, $I$ would need to contain some of the $q_i$ for $1 \le i \le n$. Or am I wrong by saying this?
The key is that you're only trying to exclude finitely many $i$. For instance, we can take $$ \epsilon = \min_{k = 1, \dots,n} |q_k - x_0| $$ then it suffices to take the interval $$ I = (x_0 - \epsilon, x_0 + \epsilon) $$ To put it another way: there is an interval excluding each single element $x_k$. It then suffices to take the intersection of these intervals, noting that the intersection of finitely many intervals containing $x_0$ will yield another interval containing $x_0$.