Why $\hat{\mathbb{C}}\setminus K$ connected $\implies {\mathbb{C}}\setminus K$ connected? ($K $ compact)

Question

Let $\hat{\mathbb{C}}=\mathbb{C}\cup \{\infty\}$ denote the extended complex plane, with the usual topology.That is $U$ such that $U$ is open in $\mathbb{C}$ and the neighbourhoods of $\{\infty\}$ $U_{\infty}=\mathbb{C} \setminus K \cup \{\infty\}$ with $K$ a compact subset of the complex plane.

We need to show that if $K$ is a compact subset of $\mathbb{C}$ then $\hat{\mathbb{C}}\setminus K$ connected $\implies {\mathbb{C}}\setminus K$ connected

I tried to prove it assuming a clopen non-trivial subset of $ {\mathbb{C}}\setminus K$ . Showing that this set is open is easy but how can i show that is also closed in $\hat{\mathbb{C}}\setminus K$?

Answer 1

Since $K$ is compact, it is closed and bounded. There is thus a ball $B$ of some radius $r<\infty$ covering $K$. Since $\Bbb C\setminus B$ is connected, $\Bbb C\setminus K$ has exactly one unbounded component.
Let $U$ be a clopen subset of $\Bbb C\setminus K$ containing the unbounded component. Then $U$ is open in $\Bbb C$ and $\Bbb C\setminus U$ is compact, hence $U\cup\{\infty\}$ is open in $\hat{\Bbb C}$. But it's also closed in $\hat{\Bbb C}\setminus K$ since $(\hat{\Bbb C}\setminus K)\setminus(U\cup\{\infty\}) = {\Bbb C}\setminus (K\cup U)$ is open.

Answer 2

Suppose WLOG that $0 \in K$, and apply the homeomorphism $z \mapsto \frac{1}{z}$ of $\hat{\Bbb{C}}$. Then $\hat{\Bbb{C}} \setminus K$ is mapped into some bounded open set $D \subset \Bbb{C}$ containing $0$. $D$ is path connected because it is connected. Can you show that $D \setminus \{0 \}$ is path connected as well (paths do not need to pass through $0$, because $D$ is open, so "there is enough space" to go around $0$ and avoid it)? Then conclude.