This is an exercise I found in a calculus book (which deals with general topology too):
Let $\Delta$ be a denumerable set. What I want to do is to build a monotone function $f: \Bbb{R} \rightarrow \Bbb{R}$ that is continuous everywhere except in the points of $\Delta$.
So let's take $\{\epsilon_j\}_{j \in \Bbb{N}}$ such that $\forall j \in \Bbb{N} : \epsilon_j \in (0,+ \infty) \subseteq \Bbb{R}$ and that the serie $\sum_{k=0}^{+ \infty} \epsilon_k$ converges. Now take a bijective function $\underset{j \; \longmapsto \;x_j}{\sigma: \Bbb{N} \rightarrow \Delta}$ (which exists because $\Delta$ is denumerable and define $$f(x) = \sum_{\{j \in \Bbb{N}:x_j = \sigma(j) \lt x\}} \epsilon_j \qquad \qquad \forall x \in \Bbb{R}$$
Now I want to prove that this function is continuous everywhere except in the points in $\Delta$, but I'm facing a problem: suppose that $\Delta$ is dense in $\Bbb{R}$ (and it could be, just take $\Delta = \Bbb{Q}$). For each point $x_0 \in \Bbb{R} \setminus \Delta$ if I consider all the possible neighbourhoods of that point in the form $(x_0 - \delta, x_0 + \delta)$ I find there at least one discontinuity of $f$, so how come I can choose a $V(f(x_0))$ neighbourhood of $f(x_0)$, when $x_0 \in \Bbb{R} \setminus \Delta$, such that $\exists U(x_0) : f(U(c) \cap dom f) \subseteq V(f(x_0))$ with $U(x_0)$ neighbourhood of $x_0 \in \Bbb{R} \setminus \Delta$?
It seems to me that this function, in the case $\Delta$ is dense in $\Bbb{R}$ is continuous nowhere.
Am I right?
No, this function is continuous on $\Delta^C$. You can have discontinuities in any neighborhood of $x \notin \Delta$, but this discontinuities are small and becomes smaller if the neighborhood shrinks. Let $x \in \Delta^c$ be fixed. For any $\varepsilon >0$ take $N \in \mathbb{N}$ so large that $\sum_{k=N+1}^\infty \epsilon_k < \varepsilon$. Then for any $|x-y| < \delta$ with $\delta := \min_{j=1,\ldots,N} \{|x-\sigma(j)|\}$. $\delta$ is not zero, because $x \notin \Delta$. We also have that, if $\sigma(j) < x$ for some $j=1,\ldots,N$, then also $\sigma(j) < y$. And similar $\sigma(j) > x$ for some $j=1,\ldots,N$ gives $\sigma(j) > y$.
For any $|x-y| < \delta'$ we get that the first $N$ terms in the sum are equal. Thus $$|f(x)-f(y)| \le \sum_{k=N+1}^\infty \epsilon_k < \varepsilon.$$
Note that you function is a limes of continuous of continuous functions. Just take finite sum and approximate the jumps by a linear function which slope increases with the number of summands. Thus this function is of first Baire class. For this functions the set of discontinuity is always of first category. In particular, the set of continuity is dense.