Covering a compact subset by a tubular neighborhood of a smooth curve

Question

Let $K$ be a compact subset of the open and connected set $\Omega\subseteq\mathbb{R}^n$, $B_\varepsilon$ the ball centered at the origin of radius $\varepsilon$. Is it true the intuitive fact that there exists an $\varepsilon>0$ and a smooth curve $\gamma_K:I\to\mathbb{R}^n$ such that $K\subseteq\gamma_K+B_\varepsilon\subseteq\Omega$?

Answer 1

Here's a quick answer, but you should make sure that everything I'm saying is justified.

First, since $K$ is compact, there is a finite set of points $\{p_i\}_{i=1}^N$ such that $K$ is covered by the balls $B_{\epsilon/2}(p_i)$.

Second, we can construct a piecewise-linear curve $\gamma$ by connecting $p_1$ to $p_2$ with a line, $p_2$ to $p_3$, and so on.

Third, you can always find a smooth, embedded curve $\gamma_K$ within a distance of $\epsilon/2$ of $\gamma$.

Therefore, any point in $K$ is within a distance of $\epsilon$ of $\gamma_K$, so $$K \subseteq \gamma_K + B_\epsilon$$ as desired.

This construction can be modified without much effort to make sure that $\gamma_K + B_\epsilon$ lies inside this set $\Omega$ as well (Hint: since $K$ is compact, the distance between $K$ and the boundary of $\Omega$ is bounded away from zero).