Htykuut

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.