Is the Alexandroff Double Circle first-countable?

Question

The Alexandroff Double Circle is the topological space with underlying set $C = C_1 \cup C_2$, where $C_i$ is the circle of radius $i$ and centre $0$ in the complex plane. The 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 $C_1$ (i.e., a union of open arcs on the circle) and $F$ is a finite subset of $C_2$.

Upon analyzing this space, I determined it is Hausdorff and compact (since $C_1$ has the normal topology and is compact, we can construct a finite subcover).

Now, I am wondering if this space is first-countable. It is clear that any point from the outer circle has a countable neighbourhood base. What about the points from the inner circle?

Answer 1

Yes, this space is first countable. Note that as each point in $C_2$ is isolated, we need only work to show that each point in $C_1$ has a countable neighborhood base. So fix some $x \in C_1$. For each $\varepsilon > 0$ by $C_\epsilon (x)$ denote the set of all $y \in C_1$ such that the shortest arc between $x$ and $y$ along $C_1$ has length strictly less than $\varepsilon$. Then for each $n > 1$ let $$U_n (x) := C_{1/n} (x) \cup \{ 2y : y \in C_{1/n}(x), y \neq x \}.$$

Clearly each $U_n(x)$ is a (basic) open neighborhood of $x$. Now given any basic open neighborhood $U$ of $x$, there is an open arc $W \subseteq C_1$ and a finite $F \subseteq W$ such that $$U = W \cup \{ 2y : y \in W, y \notin F \}.$$ Let $n$ be so large that for each $y \in F \setminus \{ x \}$ the shortest arc between $x$ and $y$ has length at least $\frac 1n$. It is then relatively straightforward to show that $U_n(x) \subseteq U$.

Answer 2

HINT: If we restrict the topology to $C_1$, we have a standard topology of torus, hence there exists countable neigbourhood basis for (restricted to $C_1$) for any point of $C_1$, but it defines the neighbourhood basis of this point on $C_1\cup C_2$.