Closed disk inside an open set
Let's fix an open set $A$ in the complex plane $\mathbb{C}$ endowed with euclidean topology.
Let's fix $z_0 \in A$ (suppose $A$ non empty).
Then obviously there exists a closed disk $\overline{B_r(z_0)}$ centered in $z_0$, of radius $r>0$ and contained in $A$.
Now, how can I show that there exists another closed disk $\overline{B_s(z_0)}$ such that $\overline{B_r(z_0)} \subseteq \overline{B_s(z_0)} \subseteq A$?
I have searched online but I can't find an answer. And I have no idea how to approach this.
Thank you!
EDIT: Actually I think I have found a way. I have to use a proper continuous function which will be minimize.
$\endgroup$3 Answers
$\begingroup$The key ingredient to use here is compactness. Suppose no such disk exists. That means that for each $\epsilon>0$, $\overline{B_{r+\epsilon}(z_0)}$ is not contained in $A$, so there exists $x\not\in A$ such that $d(x,z_0)\leq r+\epsilon$. Taking a sequence of such points as $\epsilon\to 0$, we get a sequence $(x_n)$ of points, none of which are in $A$, such that $d(x_n,z_0)$ converges to $r$. This is a bounded sequence in $\mathbb{C}$, so it has a subsequence which converges to some limit $x$. Then $d(x,z_0)=r$. Also, since each $x_n$ is not in $A$ and $A$ is open, $x\not\in A$. But now this is a contradiction since $\overline{B_r(z_0)}\subseteq A$.
$\endgroup$ $\begingroup$If $A=\Bbb C$ we are done. Now, let $A\not=\Bbb C$, so that $A$ has non-empty boundary $\partial A$. Let us consider $\text{distance}(z_0,\partial A)=\inf\big\{||z_0-x||:x\in \partial A\big\}$. From hypothesis we have $0<r<\text{distance}(z_0,\partial A)$. Now take any $s$ with $r<s<\text{distance}(z_0,\partial A)$.
$\endgroup$ $\begingroup$Hint: $A$ open implies there is $z_1\in A\setminus\overline{B_r(z_0)}$. Let $S:=\{|z-z_0|:z\in A\setminus\overline{B_r(z_0)}\} $. Now set $s:=\inf S$.
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