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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

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    $\begingroup$ If $K$ is compact and $D$ is closed in ${\mathbb R}^n$ with $K \cap D = \emptyset$, then $d(K,D) = \inf_{x \in K, y \in D}||x - y||$ is positive (try proving using sequences). So in your case let $K$ be the image of $p(t)$, let $D$ be the complement of $U$, and let $\epsilon = d(K,D)$. $\endgroup$
    – Michael Greenblatt
    Commented Dec 13, 2010 at 5:10
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    $\begingroup$ sounds like a homework problem $\endgroup$
    – Anthony Quas
    Commented Dec 13, 2010 at 5:31
  • $\begingroup$ Alternatively, the continuity of $t\mapsto \epsilon_t$ should follow from the continuity of $t\mapsto p(t)$ and the triangle inequality. I think, this question would be better suited for math.stackexchange.com. $\endgroup$
    – Alex B.
    Commented Dec 13, 2010 at 5:33
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    $\begingroup$ I confess to being surprised by the answers that have been proposed here. This question is an elementary exercise using the standard definition of compactness and how compact sets behave under continuous maps. $\endgroup$
    – Deane Yang
    Commented Dec 13, 2010 at 14:23

1 Answer 1

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$\epsilon_t = \operatorname{sup}(\{r\in (0,1) : \operatorname{B}(p(t),r)\subseteq U\}) = \operatorname{inf}(\{d(p(t),u) : u\in U\}) = d(p(t),U)$

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

$x\mapsto d(x,U)$ 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

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  • $\begingroup$ Are you sure about the second and third equal signs in the first line of your post? $\endgroup$
    – Did
    Commented Dec 13, 2010 at 7:12
  • $\begingroup$ Now I am. (doh) $\endgroup$
    – user5810
    Commented Dec 14, 2010 at 0:38
  • $\begingroup$ Now the first and third equal signs are true but the second one isn't. At all. $\endgroup$
    – Did
    Commented Dec 14, 2010 at 6:29

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