Compact subset of an open subset of Euclidean Plane.

Question

Let $E$ be a compact subset of an open subset $U$ of the Euclidean plane. Can we ensure that there is an $r>0$ such that for each $z \in E$ , the closed disc $\bar{D} (z,r)$ is contained in U ?

Answer 1

Define $s(x) = \inf_{z\in U^c} ||x-z||$. Since $s : E \rightarrow \mathbb{R}_+$ is continuous, and $E$ is compact, we know that there exists $x_{min} \in E$ such that $x_{min} = \arg\min_{x \in E} s(x)$.

Suppose $s(x_{min}) = 0$. Because $U$ is open, $U^c$ is closed. Therefore, it can easily be verified that $x_{min}$ must belong to $U^c$, which is impossible as $x_m \in E \subset U$. Indeed, if $s(x_{min}) = 0$, one can find $(x_n)_n$ a sequence in $U^c$ such that $||x_n - x_{min}|| \rightarrow_n 0$. Since $U^c$ is closed, the limit $x_{min}$ of the converging sequence belongs to $U^c$.

Therefore $s(x_{min}) = \epsilon > 0$. Then given any point $x$ in $E$, one has that $D(x, \frac{\epsilon}{2}) \subset U$, which yields the result by setting $r = \frac{\epsilon}{2}$

Answer 2

Suppose such $r$ does not exist. For every $n>0$ there exists $x_n$ in $E$ such that the closed ball $B(x_n,1/n)$ is not in $U$. You can extract a converging subsequence $(x_{n_i})$ to $x$ in $E$. There exists $r$ such that $B(x,r)$ is in $U$. There exists $N$such that $n_i>N$ implies that $x_{n_i}$,in $B(x,r/2)$. if $1/n_i<r/2$, we deduce that $B(x_i,1/n_i)\subset B(x,r)\subset U$. Contradiction.

Answer 3

In the non-trivial case $E\ne \phi$: Each $p\in E$ belongs to the open set $U,$ so take $r_p>0$ such that the open ball $B(p,r_p)\subset U.$

Now $\{B(p,\frac {1}{2}r_p):p\in E\}$ is an open cover of $E,$ so let $S$ be a non-empty finite subset of $E$ such that $\cup \{B(p,\frac {1}{2}r_p):p\in S\}\supset E.$

Let $e=\min \{\frac {1}{2}r_p:p\in S\}.$

For any $q\in E$ there exists $p\in S$ such that $q\in B(p,\frac {1}{2}r_p).$ Now for any $q'\in \overline {B(q,e)}$ we have (with $d$ denoting the Euclidean metric) $d(q,q')\leq e$ so $$d(p,q')\leq d(p,q)+d(q,q')<\frac {1}{2}r_p+e\leq \frac {1}{2}r_p+\frac {1}{2}r_p=r_p.$$ So $q'\in \overline {B(q,e)}\implies q'\in B(p,r_p)\subset U.$ That is, $\overline {B(q,e)}\subset U$ for any $q\in E.$