Is the Alexandroff double circle separable?

Question

Is the Alexandroff double circle separable (i.e. has a countable dense subset)?

The Alexandroff double circle is the 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 $z \in \mathbb{C}$ 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) and $F$ is a finite subset of $C_2$.

So far, I was able to show it is compact, Hausdorff, first countable but not second countable. Is there a way to connect this information to determining if it is separable or not?

score 0 · Answer 1 · answered Aug 06 '15 at 02:39

0

No, it's not separable. The points of $C_2$ form an uncountably infinite discrete subspace. If there were some countable dense subset $S$, there would have to be an element of $S$ in each open set $\lbrace z \rbrace$ for all $z \in C_2$. Therefore, $C_2 \subseteq S$, which contradicts $S$ being countable.

answered Aug 06 '15 at 02:39

Theo Bendit

53,568

2

It's not really that $C_2$ is an uncountable discrete subspace (the Niemytzki/Moore plane has an uncountable discrete subspace, but is still separable), but that every point in $C_2$ is isolated (i.e., the singleton ${x}$ is open). – user642796 Aug 06 '15 at 04:06
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Theo Bendit should have said :An uncountable discrete OPEN subspace. – DanielWainfleet Aug 06 '15 at 05:27

Is the Alexandroff double circle separable?

1 Answers1