Let $U\subset \mathbb{R}^m$ be nonempty, open and convex. If $[x,y]:=\{x+t(y-x)\mid t\in[0,1]\}$, the geodesic distance on $U$ is defined as: \begin{align*} d_U:U\times U &\to \mathbb{R}\\ (a,b)&\mapsto \inf\left\{\sum_{i=1}^{k-1}|x_{i+1}-x_i|: [x_i,x_{i+1}]\subset U\text{ for all }i\text{ with }x_1=a, x_k=b\right\} \end{align*} Let $D\subset U$ be a closed, enumerable subset. Defining the geodesic distance on $V:=U\setminus D$ in the same manner, prove that $d_V(x,y)=|x-y|$ for all $x,y\in V$.
The result feels intuitive, but I'm having trouble formalizing it.
Since the set $[x,y]\subset U$ is non-enumerable, I can find points $x=x_1,...,x_k=y\in[x,y]\cap V$ with $|x_{i+1}-x_i|$ arbitrarily small. I'm trying to argue that the sum $\sum_{i=1}^{k-1}|x_{i+1}-x_i|$ is arbitrarily close to $|x-y|$ as well as to $d_V(x,y)$, so it must converge to $d_V(x,y)=|x-y|$.
Does this strategy work? Any suggestions?
(1) $D=\{x_i\}$ and $D_\alpha=\{x_1,\cdots, x_\alpha\}$
For $D_\alpha$ we do the following : For $x,\ y\in U$, for $\epsilon$, there is $x_i$ s.t. $d_U(x,y) +\epsilon > \sum_{i=1}^k\ |x_i-x_{i+1}|$.
Hence there is $y_i\in U-D_\alpha$ s.t. $|x_i-y_i|<\frac{\epsilon}{2^i}$. Hence $$ \sum_i\ |y_i-y_{i+1}| <\sum_i\ |x_i-x_{i+1}| + 2\epsilon <d_U(x,y) + 3\epsilon $$
Hence we have sequence of piecewise line segment $c_\alpha := \bigcup_i\ [ y_iy_{i+1}]$
(2) We do a limit for $c_\alpha$.