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I recently encountered the Alexandroff double circle. 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. 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 $C_2$.

I want to know if the space is compact and Hausdorff.

  • If this space is compact, then every cover will have finite subcover. I'm a little lost on this. Can anyone provide a hint?

  • I believe it is Hausdorff because every pair of points containing two points on the outer circle will have disjoint open sets because one point sets on the outer circle are open. If a point $x$ is in the inner circle and a point $y$ on the outer circle, we can set $F = \{y\}$, and disjoint open sets will exist for $x$ and $y$. If we have to points on the inner circle, non-intersecting arcs on the inner circle will generate disjoint open sets. Am I correct in deducing this space is Hausdorff?

Martin Sleziak
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  • Yes, it is correct that the space is Hausdorff. Since the subspace $C_1$ has its usual topology, you know that every open cover of $C$ contains a finite family that covers $C_1$, and then ... – Daniel Fischer Jul 01 '15 at 08:41
  • and then...we can use that finite family to construct a finite subcover of C. It contacts analogous open arcs in $C_2$ minus finite subsets of $C_2$ which we can fill in by appending a finite amount of open sets from our cover. Thanks Daniel Fischer! – ThinkConnect Jul 01 '15 at 09:04

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