A question about how completely a rectifiable arc can fill a non-empty compact continuum that is the closure of its interior

Question

Let P be the Euclidean plane and let C be a compact and convex subset of P whose interior is non-empty. Does there always exist a strictly increasing sequence of positive real numbers l(1),l(2),...,l(n),... as well as a strictly decreasing sequence of positive real numbers e(1),e(2),...,e(n),... converging to zero such that, for each positive integer i, there is a subset s(i) of C which is a rectifiable arc whose length does not exceed l(i) and whose distance from each point of C is not greater than e(i)? If the answer to this question is "Yes", must the infinite sequence l(1),l(2),...,l(n),... always be unbounded? .........It looks to me intuitively that the answers to these questions should be "Yes" and that the proof should be easy. But all my attempted proofs seem to have gaps, so maybe I am missing something. I specialized the problem here to make it easier to "visualize". P could be a higher dimensional Euclidean space and C does not have to be convex.

Answer 1

No, the lengths must go to infinity.

If $s$ is a path of length $l$ in the plane and $\varepsilon>0$, then the area of the $\varepsilon$-neighborhood of $s$ is no greater than $20\varepsilon(l+\varepsilon)$. Indeed, $s$ can be divided into at most $l\varepsilon^{-1}+1$ subintervals of length at most $\varepsilon$. Pick a point on each subinterval and consider balls of radius $2\varepsilon$ centered at these points. These balls cover the $\varepsilon$-neighborhood of $s$ and the sum of their areas is at most $4\pi\varepsilon^2(l\varepsilon^{-1}+1)=4\pi\varepsilon(l+\varepsilon)<20\varepsilon(l+\varepsilon)$.

Since your $C$ has nonempty interior, it has positive area $A$. Since $e(i)$-neighborhood of $s(i)$ covers $C$, the above inequality implies that $20 e(i)(l(i)+e(i))\ge A>0$. Since $e(i)\to 0$, it follows that $l(i)\to\infty$.