I'm trying to prove a theorem(I'm not sure whether it's right)
$X\cup\{\infty\}$ is connected in $\overline{C}$(expanded complex plane) if and only if all connected components of $X$ in $C$ are unbounded.
This problem arises because the definition of simple connected domain in (Complex analysis)(Ahlfors L.V.)
I'd proved that if all connected components of $X$ is unbounded then $X\cup\{\infty\}$ is connected in $\overline{C}$,but when I try to prove another, I have no idea.
I would be very grateful if someone could answer this question.
The converse fails.
Let $A_1,A_2,A_3,...$ be the subsets of $\mathbb{C}$ given by $$ A_n = \left\{ re^{i\theta} {\,{\large{\mid}}\,} r\ge{\small{\frac{1}{n}}} \;\text{and}\; \theta={\small{\frac{\pi}{2n}}} \right\} $$ and let $X$ be the subset of $\mathbb{C}$ given by $$ X = \{0\} \,{\large{\cup}} \left( \, {\small{\bigcup_{n=1}^\infty}} A_n \right) $$ Claim:$\;X\,\cup\{\infty\}$, with the topology inherited from $\overline{\mathbb{C}}$, is connected, whereas $X$, with the topology inherited from $\mathbb{C}$, is not connected, and has the bounded set $\{0\}$ as a connected component.
Proof:
First we show that $X\,\cup\{\infty\}$ is connected in $\overline{\mathbb{C}}$.
For each positive integer $n$, let $B_n=A_n\cup\{\infty\}$.
Noting that $A_n$ is connected in $\overline{\mathbb{C}}$, and $B_n$ is the closure of $A_n$ in $\overline{\mathbb{C}}$, it follows that $B_n$ is connected.
Let $Y=\bigcup_{n=1}^\infty B_n$.
Since the sets $B_n$ share a common point, it follows that $Y$ is connected.
Then the closure of $Y$ is also connected.
But the closure of $Y$ is just $Y\,\cup\{0\}$ which equals $X\,\cup\{\infty\}$, hence $X\,\cup\{\infty\}$ is connected in $\overline{\mathbb{C}}$.
Next we show that $X$ is not connected in $\mathbb{C}$, and has the set $\{0\}$ as a connected component.
For each $n$, the identity $$ X = A_n \cup \bigl( X{\setminus}A_n \bigr) $$ yields a partition of $X$ into two closed sets, hence $X$ is not connected.
Moreover, since each $A_n$ is connected, we get, from the above partition, that each $A_n$ is a connected component of $X$.
Finally, since the connected components of $X$ partition $X$, it follows that $\{0\}$ is a connected component of $X$.