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I am trying to understand the proof of Lemma 45.1 in Bartle, The Elements of Real Analysis. The lemma is used to prove that diffeomorphisms map sets with zero content into sets with zero content. Bartle leaves to reader (with hint that is $\bar{A}$ is compact):

If $\bar{A} \subset \Omega \subset \mathbb{R}^p$ where $\Omega$ is an open set and $\bar{A}$ is closed and bounded, then $\inf \{ \|a-x\|: a\in \bar{A}, x \notin \Omega\}> 0$.

My attempt: I know $\{ \|a-x\|: a\in \bar{A}, x \notin \Omega\} = \{ \|a-x\|: (a,x) \in \bar{A} \times \Omega^c\}$ where $\Omega^c = \mathbb{R}^p - \Omega$, and the Euclidean norm $\|a-x\|$ is a continuous function of its arguments. If I knew that $\bar{A} \times \Omega^c$ was compact then I would have the infimum equal to $\|a'-x'\|$ for some point $(a',x')$. But since $\bar{A} \cap \Omega^c = \emptyset$ we must have $\|a'-x'\| > 0$.

But I am stuck when $\Omega^c$ is not compact.

WoodWorker
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  • Are these strict subsets.s – Jake Freeman Jan 11 '22 at 00:38
  • Note that A is closed and is a subset $\Omega$ which is open. – WoodWorker Jan 11 '22 at 00:39
  • Ok consider that any open set is countable intervals – Jake Freeman Jan 11 '22 at 00:48
  • Note that $\overline A$ is compact and $\Bbb R^p \setminus \Omega$ is closed. In any metric space, the distance between a compact set and a disjoint closed set is strictly greater than zero. That's because for each $n \in \Bbb N, U_n= { x \in \Bbb R^p \mid d(x, \Bbb R^p \setminus \Omega) \gt \frac 1n }$ is open so the collection of $U_n$ is an open cover of $\overline A$, which must have a finite subcover. – Robert Shore Jan 11 '22 at 00:59

3 Answers3

2

Let $\overline A=X$, and make a map as $f:X\to\Bbb R$, $f(x)=\inf\{\|x-a\|\mid a\in\Omega^c\}=d(x,\Omega^c)$. Since $x\in\Omega$, there exists $\epsilon_x>0$ such that $\|x-p\|<\epsilon_x\Rightarrow p\in\Omega$. This shows $f(x)\ge\epsilon_x$.

Since $X$ is compact and $f$ is continuous, $fX$ is also compact. So there exists $\epsilon>0$ such that $fX\subseteq[\epsilon,\infty)$. This means $d(X,\Omega^c)=\inf fX\ge\epsilon>0$.

MH.Lee
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2

The key here is to work with the function $a \mapsto d(a, \Omega^c) := \inf\{\|a-x\|:x \in \Omega^c\}$, where it is easy to prove both that

$$\tag{1}\inf\{\|a-x\|: a \in \bar{A}, x \in \Omega^c\}= \inf\{d(a, \Omega^c): a \in \bar{A}\},$$

$$\tag{2} d(\cdot,\Omega^c) \in C(\bar{A})$$

Since $\Omega$ is an open set, for any $a \in \bar{A} \subset \Omega$ there exists $\delta_a > 0$ such that the open ball $B(a;\delta_a)$ is contained in $\Omega$. If $x \in \Omega^c$, then $x \notin B(a; \delta_a)$ and $\|a-x\| \geqslant \delta_a > 0$. This implies that $d(a, \Omega^c) \geqslant \delta_a > 0$.

Since $\bar{A}$ is closed and bounded and, hence, compact and the distance function is continuous, there exists $a^* \in \bar{A}$ such that

$$\inf\{\|a-x\|: a \in \bar{A}, x \in \Omega^c\}=\inf\{d(a, \Omega^c): a \in \bar{A}\}= d(a^*,\Omega^c) \geqslant \delta_{a^*} > 0$$

Proof of (2). To prove that $a \mapsto d(a,\Omega^c)$ is continuous note that for $a_1,a_2 \in \bar{A}$ and any $x \in \Omega^c$, we have, by the reverse triangle inequality

$$d(a_1,\Omega^c)- \|a_1-a_2\| \leqslant \|a_1-x\| - \|a_1-a_2\| \leqslant\|a_2-x\|$$

Thus, $d(a_1,\Omega^c)- \|a_1-a_2\| \leqslant \inf \{\|a_2 - x\|: x \in \Omega^c\} = d(a_2,\Omega^c)$ and after rearranging,

$$d(a_1,\Omega^c) - d(a_2, \Omega^c) \leqslant \|a_1 - a_2\|$$

Whence, switching $a_1$ and $a_2$ leads to

$$|d(a_1,\Omega^c) - d(a_2, \Omega^c)| \leqslant \|a_1 - a_2\|,$$

and it follows that $a \mapsto d(a, \Omega^c)$ is continuous.

RRL
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1

To solve it your way, choose $M$ such that $\Omega_M = \{ y \in \Omega^c \mid d(X, y) \leq M \} \neq \varnothing.$ Then $\Omega_M$ is compact (because it's closed and bounded) and $\inf \{d(x, y) \mid x \in X, y \in \Omega_M \} = \inf \{ d(x, y) \mid x \in X, y \in \Omega^c \}$, so consider $X \times \Omega_M$.

Robert Shore
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