What do open sets look like in the Alexandroff double cirlce?

Question

I recently encountered the following topological space, called the Alexandroff double cirlce:

The underlying set is $C = C_1 \cup C_2$, where $C_i$ is the circle of radius $i$ and centre $0$ in the complex plane. Basic open sets are:

• $\{ z\}$ for every $z$ with $|z| = 2$, and

• $U \cup \{ 2z : z \in U\} \setminus F$, where $U$ is open in the normal topology of $C_1$ and $F$ is a finite subset of $C_2$.

Just to make sure I understand this space, I have two questions.

1. Does $U \cup \{ 2z : z \in U\} \setminus F$ refer to all points in $U \cup \{ 2z : z \in U\}$ except those in $F$?
2. When it says $U$ is open in the normal topology of $C_1$, what does this mean? Are they just epsilon balls in the complex plane?

Answer 1

Yes, $U∪\{2z:z∈U\} \setminus F$ is all points in $U∪\{2z:z∈U\}$ except those in $F$.

The open sets of $C_1$ would be unions of open 'arcs' on the unit circle. I'm taking 'circle' to mean the set of points $\{ z \in \mathbb{C} : |z| = 1 \}$ rather than the 'disc' $\{ z \in \mathbb{C} : |z| \leq 1 \}$.