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I'm hoping someone can comment on if my logic on the Alexandroff double cirlce not being second countable is right.

The Alexandroff double circle has underlying set $C = C_1 \cup C_2$ where $C_i = \{ z \in \mathbb{C} : |z| = i \}$ is the circle of radius $i$ centred at the origin in the complex plane. The basic open sets are

• $\{ z\}$ for every $z \in C_2$, and

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


I believe this space is not second-countable for the following reason. For every element $z$ on the outer circle, the singleton set containing it is open. Thus if it had a countable basis, it would have to contain the set of all singleton sets $\{z\}$ on the outer circle. But that would make it uncountable.

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  • $\begingroup$ Your reasoning is correct. $\endgroup$ – Brian M. Scott Sep 1 '15 at 2:12

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