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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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closed as too localized by Andrey Rekalo, Yemon Choi, fedja, Deane Yang, Andres Caicedo Dec 13 '10 at 6:03

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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 $\epsilon = d(K,D)$. –  Michael Greenblatt Dec 13 '10 at 5:10
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sounds like a homework problem –  Anthony Quas Dec 13 '10 at 5:31
    
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. –  Alex B. Dec 13 '10 at 5:33
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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. –  Deane Yang Dec 13 '10 at 14:23
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1 Answer 1

up vote 1 down vote accepted

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

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

$x\mapsto d(x,U)$ 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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Are you sure about the second and third equal signs in the first line of your post? –  Did Dec 13 '10 at 7:12
    
Now I am. (doh) –  Ricky Demer Dec 14 '10 at 0:38
    
Now the first and third equal signs are true but the second one isn't. At all. –  Did Dec 14 '10 at 6:29
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