Im trying to prove that a finite number of compact and rectifiable curves cannot cover any open disk in $\Bbb R^2$. I dont know if this is true but it seems.
If a curve is a $1$-dimensional submanifold then the result is clear because the dimension of a submanifold is unique, but I dont know how to approach this problem for any kind of compact curves.
Can someone show me if this is false or true? Preferably with the most elementary methods (no measure theory, by example).
I found a very "simple" answer using elementary geometry.
Observe that for any path that connect $m$ points there is a minimum distance between the connection of any two different points. Hence the length of any path that connect $m$ points is bounded below by $L:=\ell\cdot (m-1)$, where $\ell$ is the minimum distance between all pairs of different points and $m-1$ is the minimum number of segments required to connect all points.
Now consider a grid of $m$ points equally spaced in the plane by a distance $\ell$, by example the vertex of a tessellation of the plane by isosceles triangles.
Then define a sequence of finer grids as follows: first consider the vertex of a isosceles triangle of side $\ell$, this is the grid $G_0$. Now defines recursively $G_{n+1}$ by adding to $G_n$ the points in the middle of any side of an isosceles triangle defined by the points of $G_n$. Then if the minimum distance between points in $G_0$ is $\ell$ then the minimum distance between points in $G_n$ is $\ell/2^n$.
The number of points in $G_n$ can be counted using the triangular number $T_n:=\binom{n+1}2$ for $n$ vertex in one side of the starting triangle (that is, the number of points in the segment that join any two points in $G_0$). The number of vertex $a_n$ in one side of the starting triangle is defined by the recurrence relation $a_0:=2$ and $a_{n+1}=2 a_n-1$. A brief calculation shows that $a_n=1+2^n$.
Thus the length of any path that connect the grid $G_n$ is bounded below by $$ L_n=\frac{\ell}{2^n}\cdot (T_{a_n}-1)=\frac{\ell}{2^n}\cdot\left(\frac{(a_n+1)a_n}{2}-1\right)\\=\ell\cdot\left(\frac{(2+2^n)(1+2^n)}{2^{n+1}}-\frac1{2^n}\right)=\ell\cdot(2^{n-1}+3/2) $$ and because $\lim_{n\to\infty}L_n=\infty$ then we can conclude there is no possible that a rectifiable curve cover an open disk.