Suppose that $A$ is a given subset of $I=[0,1],\ $ and
$ \left\{ x_j = \frac{j}{m} \right\}_{j=0}^{m}\ $ is the $m$-partition of $I$, and $\nu(m)$ is the number of
$\ [x_{i-1},x_{i}]\ $ such that $\ [x_{i-1},x_{i}]\cap A \neq \emptyset\ $ for intervals $\ [x_{i-1},x_{i}]\ $ belonging to the $m$-partition. Also assume that $M>0$ is such that

$$ \forall_{m>M}\quad \nu(m) \ge \ln(m) $$

Does it follow that $A$ is dense on some open non-empty subinterval of $I$?

>***Remark** (editorial by wlod): the rest of the post is unclear to editor wlod. Once the things get clear, this remark should be removed. ### This remark is here to avoid a prolonged (if momentary) discussion in comments. ### Also, Wlod followed by Andreas have provided counter-examples, and -- to make it better -- they need not the complication of constant $M$. Wlod stated the result for general functions $f$ in place of $\ f:=\ln$. Then Andreas had it for a wider class of functions $f$ (but Wlod's proof handles it too).*

An example:

Let $A = \{\frac{1}{j}\}_{j=1}^{\infty}$. By some calculation we can get the value of $n$:
$$1,7,23,70,211,649,$$
when $m$ is $1,10,100,1000,10000,100000$,respectively.

Notes: I want to emphasize that you cannot construct $A$ with $m$ related to the precondition, because $m$ is given and changing to $\infty$ after $A$ has been supposed. Even if you want to construct $A$ with some variables, please notice that when $A$ is constructed, your variables are fixed and cannot follow the change of $m$.