About Hole of compact set in $\mathbb{C}$

Question

Given a compact set $K$ of complex plane $\mathbb{C}$. A hole of $K$ is defined as bounded components of its complement.Then decomposing $$\mathbb{C}/K =\Omega_\infty \cup \Omega_1 \cup\Omega_2\cup.....$$

where $\Omega_\infty$ is unbounded component and $\Omega_i$ are holes .Then it say there will be only finite number of holes or none at all.

My doubt is why there would be a unbounded component and why there are only finite number of holes.

And under what condition there will be no hole at all.

thanks in advanced.

Answer 1

Let $C_{n}$, $n>0$, be the circle of center $(0,\frac{1}{n})$ and radus $\frac{1}{n}$. And $$K=\bigcup_{n>0}C_{n}$$ Then $K$ is a compact in the complex plan and \ $\mathbb{C}$\ $K$ has infinite number of holes.