I'm looking the proof to show that a set $S$ is open iff each point in $S$ is an interior point.
Def) A set is open if it does not contain any of its boundary points
Def) A point $z$ is said to be a boundary point of set $S$ if point $z$ is neither interior point nor exterior point
Def) A point $z$ is said to be an exterior point of a set $S$ when there exists a neighborhood of it containing no points of $S$
Def) A point $z$ is said to be an interior point of a set $S$ whenever there is some neighborhood of $z$ that contains only points of set $S$
The proof goes on like this.
'Let $G$ be an open region in $\mathbb C$ and $dG$ denote it's boundary. Let $a \in G$ be any point in $G$. Choose $\epsilon = \min\{|a-b|:b \in dG\} $ then region $|z-a|<\epsilon$ is contained in $G$. So $a$ is interior point of $G$ and as $a$ was chosen arbitrary, all points of $G$ must be interior point.'
But is the $\epsilon$ defined by using the min well-defined all the time? I think there might be some exceptional cases when $dG$ is infinite. So I feel a little doubtful on this proof. (I know the proof without using any 'min' but still want to figure out the validity of the above proof.)
Best Answer
Let it be that $\partial G$ is infinite.
It is our aim to show that for fixed $a\in\mathbb C$ there exists some $b\in\partial G$ such that $|a-b|\leq|a-x|$ for every $x\in\partial G$.
First note that $\partial G$ as intersection of the closed sets $\mathsf{cl}(G)$ and $\mathsf{cl}(G^{\complement})$ is closed.
Now find positive $r$ such that $F:=\{z\in\mathbb C\mid|z-a|\leq r\}\cap\partial G\neq\varnothing$.
Then $F$ is bounded and closed, hence is compact.
On $F$ we have the continuous function prescribed by $z\mapsto|a-z|$ and a continuous function on a compact set takes a minimum on that set.
Now let $b\in F$ such that $|a-b|$ is that minimum.
Then $b$ has the property mentioned above.