Show that there exists function $f:[0.1] \to \mathbb{R}$ such that it has no derivative in no point of a dense subset of $[0,1]$. To do that I have created a numbering of of rational expressions over $[0,1]$ in the form of
\begin{align*}
\{q_{k} | k \in \mathbb{N} \}
\end{align*}
with function
\begin{align*}
g(t) = \sum_{k: q_{k} \leq t} \dfrac{1}{2^k}.
\end{align*}
It's easy to show that $g$ is increasing. I am supposed to use that fact to define an appropriate function $f$ to finish up, but I have no idea what.
 A: More than you might care to know about this function:
The standard example of a monotonic function which is discontinuous precisely at the rationals is given by
$$f_φ(x)=∑_{φ(n)<x} \frac{1}{2^n}, \tag{1}$$
where $φ$ is a bijection from $\mathbb N$ onto $\mathbb Q$.
The poster (correctly) offered this function (essentially) as an example of an increasing function with a dense set of points of discontinuity.  In fact the set of discontinuities is exactly the rationals.  Clearly, too, the function $f_φ$ fails to have a derivative at each rational (simply because a derivative cannot exist at a point of discontinuity).
Is there more that can be said?  Never ask a mathematician that since there is always more.

*

*This function has discontinuities at each rational and is continuous at each irrational.  Can this be reversed?  Is there a function continuous at each rational and discontinuous at each irrational? This is not possible for monotonic functions: every monotonic  function has at most countably many discontinuities.  More surprising (and more difficult to prove): there is no function, monotonic or not, that is continuous at each rational number and discontinuous at each irrational number.


*Functions like this are called saltus functions [Latin: saltus=jump, leap].  This function does all of its growing on the countable set of rationals where it has a jump up--it does no growing on the irrationals.  Important class of functions of which (1) is your standard interesting example.  Every monotonic function can be expressed as a sum of a continuous monotonic function and a monotonic saltus function.  Worth studying.


*Another feature of this function is that it is continuous on the right at every point and it has a limit on the left at every  point.  At a countable set (here of  rational points) only do  the left and right limit differ.  Such functions are called   càdlàg functions (French: "continue à droite, limite à gauche").  Again an important class of functions with this one as a good illustration.  Important in stochastic processes, Fourier series, etc.


*A reasonable question in light of this posted question:  This function is discontinuous at the rationals so it has no derivative at any rational point.  But it is continuous at the irrationals.  So is it differentiable at the irrationals? The answer is tricky.  It must be differentiable at many irrational numbers because it is a monotonic function.  On the other hand (more delicate) there are lots of irrational points where it does not have a derivative.  See #5, #6, #7 for these, taken from reference [1].


*In every nontrivial interval there exists a nowhere dense null set $Z$ of irrational numbers such that every $f_φ$ has uncountably many points of nondifferentiability in $Z$. (See [1].)


*If a one-sided derivative of an $f_φ$ exists at a point, then it is either infinite or it vanishes. (See [1].)


*For every countable set $C$ there exists an $f_φ$ whose derivative exists (and vanishes) at every point of the set $C$.  (See [1].)

REFERENCE:
[1]  Kuba, Gerald. On the differentiability of certain saltus functions.
Colloq. Math. 125 (2011), no. 1, 15–30.
