I'm in trouble to prove this:
Let $G$ be a region such that $a \in G$. Prove that $G \backslash \{a\}$ is a region.
MY ATTEMPT: $G \backslash \{a\}$ is open as it is nothing but $G \cap \{a\}^c$; intersection of two open sets. I have tried to prove connectedness from contradiction. We assume it is not connected, i.e. there exists two open sets $U$ and $V $ such that $G$ \ {$a$} $=$ $U \cup V$ and $U \cap V = \emptyset$.
Now I claim that there exists an open ball, say $B(a,r)$, contained in either $U \cup \{a\}$ or in $V \cup \{a\}$. If my claim is true then $G =U_1\cup V_1$, with $U_1 = U \cup \{a\}$ or $U_1 =U$. Similarly I take $V_1 = V\cup \{a\}$ or $V$. If $B(a,r)$ is contained in $V \cup \{a\}$ i rename it $V_1$ and choose $U_1 = U$ and similarly if $B(a,r) \subset U \cup \{p\}$ , I rename it $U_1$ and take $V_1 =V$.
Clearly $U_1 \cap V_1 = \emptyset$, which draws a contradiction that $G$ is not connected i.e. not a region.
My problem is HOW TO PROVE MY CLAIM that there exists such an open ball $B(a,r) \subset U \cup \{a\}$. I have thought about an intuitive approch; as $G$ is open there exists an open ball $B(a,r') \subset G =U\cup V \cup \{a\}$. If $B(a,r') \subset$ one of $U \cup \{a\}$ or $V \cup \{a\}$ , I take $r' = r$ and my claim is proved and I'm happy. If that open ball is not contained in any of those two I assume that there is no radius $r$ for which an open ball containing $a$ is a subset of $U \cup \{a\}$, so my intution says that if I repeatedly decrease the radius after a certain amount of time that open ball will be contained in $G \backslash U = V \backslash \{a\}$. So again my claim is proved.
But I think that I need to be more rigorous, which I'm not doing here. How can I do it? Or is my intution is wrong? Thanks for reading.
A region is an open and connected subset of $\mathbb C$.
Clearly, $\mathbb C\setminus \{a\}$ is open since it is the complement of the single point $\{a\}$ and single points are closed in Hausdorff spaces.
Note that, removing one point from the plane doesn't make the plane disconnected. In fact it is still path-connected. Say, $a,b$ and $c$ lie on the same line. Then you can go from $b$ to $c$ by drawing a straight line from $b$ to $d$ and a line from $d$ to $c$ where $d$ is any point in the plane different than $a$.
As a matter of fact, removing a countable set from the plane doesn't effect path-connectedness.