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Say $U \subset \mathbb{R}^n$ is open and $X \subset U$ is compact. I want to find a number $\epsilon > 0$ such that for all $x \in X$, $B_\epsilon(x) \subset U$. My plan is to define a function $\rho : X \to [0.\infty)$, where $\rho(x) = \sup\{\epsilon \in (0,1] : B_\epsilon(x) \subset U\}$. If this function is continuous, we can find its minimum because $X$ is compact, and then we're set. But is this function actually continuous?

Nick A.
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  • Is $U$ itself an open ball? If $U$ is not convex, there might not be an open ball based at a point in $X$ which is a strict subset of $U$ and contains all of $X$. Why do you want a ball $B_\varepsilon(x)$ and not just an open set containing $X$ that is itself a subset of $U$? –  Apr 02 '20 at 15:09
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    @Renard I'm not requiring that any single open ball contain all of $X$, just that there's a single radius such that all the balls of that radius around each point in $X$ lie inside $U$. My end goal is to show that $U$ contains an $\epsilon$-neighborhood of $X$ for some $\epsilon$. – Nick A. Apr 02 '20 at 15:18

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The continuity of the distance function $d(x,U^c)$ as a function of $x\in \mathbb R^n$ is a routine exercise. Restricting $x\in X$ only has the effect of making this function positive (because $X$ is (closed) compact and contained in (open) $U$).

The only additional ingredient to consider is the limitation of $\rho(x)$ to take values that are at most $1$. So the Reader is asked to show that:

$$ \rho(x) := \min \{1,d(x,U^c)\} $$

is accordingly a continuous function of $x\in X$.

hardmath
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