I am dealing with the same set based on my previous question. I want to show that the set $H = \{z \in \mathbb{C} : 0 < |z| < 1\}$ is NOT simply connected, using the following definitions, that our teacher insists upon us using:
A region $G$ is simply connected iff $\mathbb{C}^{\infty} \setminus G$, its complement in the extended plane, is connected in $\mathbb{C}^{\infty}$.
A set $H$ is not connected if $\exists$ open sets $U$ and $V$ such that $$U \cap H \neq \varnothing, V \cap H \neq \varnothing, \ H \subseteq U \cup V, \ \mathrm{and} \ U \cap V = \varnothing.$$
If I can show that the set $\mathbb{C}^{\infty} \setminus H$ is not connected in $\mathbb{C}^{\infty}$, then I will have that $H$ is not simply connected.
But how can I come up with sets $U,V$?
Thanks in advance for help.
Note that the punctured disk is punctured, in that it does not contain $0$. Further, you can enclose $0$ in a small open ball $B$, and $\{ z:|z| \geq 1 \}$ in the complement of a closed ball. Calling this ball $C$, we need it to be such that $B \cap C^{\complement} = \emptyset$. The complement of $H$, which is $ \{ 0\} \cup \{ z: |z| \geqslant 1 \} $, is contained in these two disjoint open sets.