I was asked the follwing question and wondered if my attempt is correct:
Let $\Omega\subset \Bbb R^n$, be an open bounded set, let $F\subset \Omega$ be a closed and non empty subset.
Define: $d=inf_{x\in F , y\in \Omega^c} ||x-y||$. Prove $d>0$
My attempt:
Notaion: $B_r(x)$ is an open ball with radius $r$ and center at $x$ while $\overline B_r(x)$ is the closed ball.
$\Omega$ is bounded hence exist $x\in \Omega$ and $r$ such that $\Omega \subset B_r(x)$, since $F\subset \Omega$ we get $F \subset \Omega \subset B_r(x)$ thus $F$ is also bounded. $F$ is closed thus $F^c$ is open by definition.
Let $y\in F^c$ since $F^c$ is an open set every point is in the interior thus exist an open ball centered at $y$ with radius $r'$ such that $B_r'(y)\subset F^c$, clearly $\forall x\in F: ||x-y||>r'$ cause if not $x\in B_r'(y)$ which contradicts $B_r'(y)\subset F^c$.
$\Omega$ is an open set thus $\Omega^c$ is closed. Cleary $\Omega^c \subset F^c$ since $F \subset \Omega$, let $y'\in \Omega^c, x'\in F$ by the two arrguments before $||x'-y'||$ is more than some $\delta>0$.
Observe $\overline B_\delta(x')$, and define $A=\overline B_\delta(x') \bigcap \Omega^c$. Cleary $A$ is bounded $\forall z\in A: ||z||<||x'||+\delta$ and its the intersection of two closed sets thus $A$ is also closed. By Heine–Borel theorem $A$ is compact.
Define $f(y_0)=||x'-y_0|| \forall y_0\in A$, $f$ is continuous on $A$ and since $A$ is compact by the Extreme value theorem $f$ gets a minimum on $A$.Let $y_1 \in A$ be that minimum.
$y_1 \in F^c$ since $A \subset \Omega^C \subset F^c$, thus by what we showed before $||y_1-x'||$ is bigger than some $k>0$ hence $f(y_1)>k>0$ now let $a \in \Omega^c-A$, cleary $a\in \Omega^c$ thus $a$ is not in $F$ hence $||a-x'||> \delta$.
We got that $||a-x'||> \delta>||y_1-x'||>k>0$ hence for an arbitrary $x'\in F$ and for arbitrary element in $\Omega^c$ the distance is always stricly bigger than $0$ thus $d>0$ as desired.
Any feedback would be appreciated!
Use the fact that $\Omega$ is open and bounded, so doesn't contain its non-empty boundary. F is closed, so contains it's boundary. Since F is a subset of $\Omega$, the boundary of F intersect boundary $\Omega$ is the empty set. This condition proves $d>0$.
$d=0 \iff \forall \epsilon>0, \exists x \in F, y \in \Omega^c, ||x-y||<\epsilon.$
Fix a point $y_0\in \Omega^c$. Define $d_{y_0}=\inf_{ x \in F} ||x-y_0||$
Since $F$ is closed $d_{y_0}=0 \iff x=y_0$. A contradiction since $x$ and $y_0$ are in disjoint sets. So $d_{y_0}>0$.
$d=0 \implies d_{y_0}=0$ so $d_{y_0}>0\implies d>0. $