The following question seems intuitively true, but I'm unable to see the proof. While I could prove it when $U=\mathbb{R}^n$, but for other open sets, I do not have a proof. Although I could construct $\alpha:[0,1]\to\mathbb{R}^n$ satisfying 1, 2 and 3, I can't force it to remain in $\bar{U}$. Could you please let me know if it's true. 


**Question:**


Let $n\geqslant 2$, $U\subset\mathbb{R}^n$ be open, bounded, connected, $C^\infty$ with $\bar{U}$ path connected and let $x\_0\in\bar{U}$. Let $(p\_n)\_{n\in\mathbb{N}}$ be a sequence in $\bar{U}$ such that $p\_n\to p$ in $\bar{U}$. Does there exist a $C^\infty$ path $\alpha:[0,1]\to \bar{U}$ and a sequence $(t\_n)$, $t$ in $(0,1)$ such that $t\_n\to t$ and


1. $\alpha(0)=x\_0$.
2. $\alpha(t\_n)=p\_n$, and
3. $\alpha(t)=p$.



Existence of Smooth path in a Domain through a Sequence of Points
In my other answer, I show how the result for $\mathbb{R}^N$ can be extended to arbitrary $U$. Here I prove that the result is actually *false* for $U = \mathbb{R}^N$ (so in a sense my other answer is completely void).


Consider $p = 0$ and $p\_n$ such that the shortest path that includes all $p\_n$ contained in the shell $S\_k = \{x \in \mathbb{R}^N : 2^{-k - 1} < |x| < 2^{-k}\}$ is $k$. This is easily achieved if $p\_n$ is defined to be the enumeration of the set $$\bigcup\_{k = 0}^\infty S\_k \cap (\varepsilon\_k \mathbb{Z})^N $$ for $\varepsilon\_k > 0$ decreasing sufficiently fast.


If $\alpha$ is a path with the desired properties, then $\alpha$ passes through all points $p\_n$ contained in $S\_k$, and so the length of $\alpha$ is not less than $k$. Since $k = 0, 1, 2, \ldots$ is arbitrary, we get a contradiction: being smooth, $\alpha$ must have finite length.


