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Let $B \subseteq E \subseteq \mathbb{R}^n$ where $B$ is compact relative to $E$, $E$ is open relative to $\mathbb{R}^n$. Prove that there exists $\epsilon > 0$ such that $A:= \{v \in \mathbb{R}^n \mid d(v,B) \leq \epsilon\} \subseteq E$.

My attempt:

I must somehow be able to pick an $\epsilon$ such that $\inf_b d(v,b) \leq \epsilon \implies v \in E$

So, I took a covering $\{B_E(b,\epsilon_b) \mid b \in B\}$ of $B$ with $B_E(b, \epsilon_b) \subseteq B$

Then, by compactness, there is a finite subcover and I set $\epsilon$ to be the minimum of the radii of the balls in the subcover, but couldn't prove that this $\epsilon$ works.

Any ideas?

  • take $\epsilon$ to be half that minimum – orangeskid Apr 09 '18 at 12:34
  • When you say "$E$ is open relative to $B$" what do you mean? As far as I know a set cannot be open with respect to a subset of itself, would $E$ open in $\mathbb{R}^d$ be correct? Also I don't see why $B_E(b, \epsilon_b)$ should lie inside $B$, I feel you mean $B_E(b, \epsilon_b) \subset E$. (or possibly something else). Bar that I would say the proof is fine if you choose $\epsilon$ to be anything strictly less than the minimum of the radii of the balls in the subcover. – S. Dewar Apr 09 '18 at 13:20
  • Take a look at https://math.stackexchange.com/questions/1929819/does-every-neighborhood-of-a-closed-set-f-in-a-metric-space-contain-an-varep – Jon Warneke Apr 09 '18 at 14:42

2 Answers2

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I think it is more convenient to work with sequentially compact (since we are in $\mathbb{R}^n$). Since $B \subset E$, the distance $d(B, E^c)$ is strictly greater than zero (since the function $d(x, E^c)$ must have a minimum over $B$ since $B$ compact. Also, it cannot be $0$). Then take half that.

mvenzin
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Let $f : B \to \mathbb{R}$ be given by $f(b) = d(b, E^c) = \inf\{d(b,x) : x \in E^c \}$, where $E^c$ is the complement of $E$ in $\mathbb{R}^n$. Show that (1) $f$ is continuous and (2) $f > 0$. Then, since $f$ is a continuous function on a compact set, by the extreme value theorem, it has a minimum, and by (2) the minimum is positive; call it $\varepsilon > 0$. Then show that (3) the $\varepsilon$-neighborhood of $B$ is contained in $E$.

Jon Warneke
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