Consider $[0, 1]$ with its Euclidean metric and the set $A ⊆ [0,1]$ of real numbers $x$ that contain only digits $0$ and $1$ in their decimal expansion. It is clear to me why this set is not dense : the open set $(0.4,0.5)$ obviously does not intersect $A$. But a more interesting question is : is it nowhere dense / the complement of a meagre set ?
I have the strong feeling that it is nowhere dense, this set makes me think a lot about the set $S=\{\frac1n \lvert n \in \Bbb N\}$ that is nowhere dense. I have a possible proof : we take for example the open set $U=[0,0.1)$ then it contains the open subset $(0.02,0.03)$ that is disjoint from $A$. Now we can take an arbitrary open set $U$ in $[0,1]$ ($U\neq (a,1]$ because it is an easy case to deal with), if it does not intersect $A$, then we are good. If it intersects it (I struggle to show the intersection has got more than $1$ element), let $a$ and $b$, $a<b$, in the intersection. $a=0.x_1x_2x_3\cdots$ and $b=y_1y_2y_3\cdots$, let $k:=\min\{n \in \Bbb N \lvert x_k \neq y_k\}$, then the open set $V=(0.x_1\cdots x_k 2,0.x_1\cdots x_k 3)$ is an open subset of $V$ that does not intersect A. So $A$ is nowhere dense.
And A is not the complement of a meagre set. Otherwise, $A^C$ would be a meagre set. But since $[0,1]$ is complete, it is a Baire space so $A^C$ that clearly has non-empty interior is not meagre.
Is this correct ? This is supposed to be an easy question but my proof for the nowhere dense part seems too long. Is there a faster way ?