Compact subset inbetween another compact subset

Question

Assume we have $A\subset \mathbb{R}^n$ open (regarding standard topology). If we have $B\subset A$ compact with dist$(B,\partial A)=\epsilon >0$, can we find $C\subset A$ compact with $B\subset C \subset A$ and dist$(B,\partial A)= \delta >0$ such that $\frac{\epsilon}{2}\leq \delta<\epsilon$? Intuitively this should be possible by simply taking $\delta\in\big[\frac{\epsilon}{2},\epsilon\big)$, but I find it difficult to argue rigorously. Thanks for any help!

Answer 1

For any $\delta\in (0,\epsilon)$, let $U=\{z\in A\mid \text{dist}(z,\partial A)>\delta\}$, which is open by the continuity of the map $x\mapsto \text{dist}(x,\partial A)$.

Then $B$ intersects $U$, so in particular $U\neq \emptyset$. Since $\mathbb R^n$ is connected, $\partial U\neq \emptyset$, so choose a point $c\in \partial U$. By continuity we have $\text{dist}(c,\partial A)=\delta$, and moreover, since $c\in \bar{U}\subseteq \bar{A}$ and $c\notin \partial A$, we have $c\in A$. Therefore let $C=B\cup \{c\}$.

Remark

The previous argument was intended to avoid using properties of $\mathbb R^n$ other than connectedness.

In $\mathbb R^n$, however, you can also directly argue that the closed $\epsilon-\delta$ neighborhood $C=\{x\in \mathbb R^n\mid \text{dist}(x,B)\leq \epsilon-\delta\}$ is certainly closed and bounded and contained in $A$ (since otherwise* we would have $\text{dist}(\partial A,B)\leq \epsilon-\delta$), and by the triangle inequality satisfies $$\text{dist}(\partial A,C)\geq \delta.$$

Moreover, by compactness we can choose $a\in \partial A$, $b\in B$ such that $\text{dist}(b,a)=\text{dist}(B,\partial A)=\epsilon$, and then there is a point $c$ on the segment joining these two with $\text{dist}(a,c)=\delta$ and $\text{dist}(b,c)=\epsilon-\delta$. Therefore $c\in C$ and so $\text{dist}(\partial A,C)=\delta$.

The purpose of arguing this way is you get an additional nice property for $C$, namely that $C$ contains an open $\epsilon-\delta$ neighborhood of $B$. Note that this argument works if we replace $\mathbb R^n$ with any proper geodesic metric space.

[*Here we tacitly use the fact that $\text{dist}(\partial A,B)=\text{dist}(\mathbb R^n\backslash A, B)$, which holds because balls in $\mathbb R^n$ are connected, and will not apply in general metric spaces, though will apply in a geodesic space.]