This may have a simple answer, but I'm not getting anywhere.


If $U$ is an open set in $\mathbb{R}^n$ (usual topology), and $p:[0,1] \to U$ is a continuous path, from $x=p(0)$ to $y=p(1)$, with $x,y \in \mathbb{R}^n$, then can we find an $\epsilon \in (0,1)$ such that $\cup\_{t \in [0,1]} B(p(t), \epsilon) \subseteq U$?


Obviously, for any given $t \in [0,1]$, there exists an $\epsilon\_t \in (0,1)$ such that $B(p(t), \epsilon\_t) \subseteq U$, because $U$ is open. Also we could also take $\epsilon\_t = \sup(\{ r \in (0,1) ; B(p(t), r) \subseteq U \})$, so then I'm tempted to take $\epsilon = \inf(\{ \epsilon\_t ; t \in [0, 1] \})$. But it's not clear to me how to prove that this last expression can't be zero. This does work if the map $t \mapsto \epsilon\_t$ is continuous, but I can't see how to prove that either.


Any help/insight would be appreciated. 


Thanks



epsilon tube around continuous path in open set in R^n
$\epsilon\_t = \operatorname{sup}(\{r\in (0,1) : \operatorname{B}(p(t),r)\subseteq U\}) = \operatorname{inf}(\{d(p(t),u) : u\in U\}) = d(p(t),U)$


$t\mapsto \epsilon\_t \; = \; t\mapsto d(p(t),U) = (x\mapsto d(x,U))\circ (t\mapsto p(t))$


$x\mapsto d(x,U)$ and $(t\mapsto p(t))$ are both continuous, so $(x\mapsto d(x,U))\circ (t\mapsto p(t))$ is also continuous. Therefore $t\mapsto \epsilon\_t$ is continuous.


QED


