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Consider the space $X = \{ z \in \mathbb{C} : |z| \leq 1 \} \cup \{ z \in \mathbb{C} : |z| = 2 \}$ with the topology generated by the following basic open sets:

  • the basic open neighbourhoods of $w \in X$ with $|w| < 1$ are just the usual (Euclidean) open neighbourhoods of $w$ in the open unit disc,
  • the basic open neighbourhood of $w \in X$ with $|w| = 2$ is the singleton $\{ w \}$, and
  • the basic open neighbourhoods of $w \in X$ with $|w| = 1$ are of the form $U \cup \{ 2z : z \in U, |z| = 1, z \neq w \}$ where $U \subseteq \{ z \in \mathbb{C} : |z| \leq 1 \}$ is open in the usual (Euclidean) topology on the closed unit disc.

Is this space $X$ compact?

The subspace $K = \{ z \in X : |z| = 1\text{ or }|z| = 2 \}$ is just the Alexandroff Double Circle, which is compact. I'm not sure if $X$ is similarly compact.

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up vote 1 down vote accepted

The resulting construction remains compact. Consider a covering of the construction by open sets $X_i$. These open sets contain neighborhoods of the form you just described. Then since the unit disk is compact, there is a finite number of them (the $X_i$'s) that cover it call them $X_1,\dots,X_n$. Each one contains at least one neighborhood of the form $U_i=W_i \cup p(V_i \setminus \lbrace w \rbrace)$. Since these cover the smaller circle, they cover, by projection, the larger circle except possibly for some finite number of points $v_i$ on the larger circle. But these were already covered by a finite number of the $X_i$'s (maybe different than $X_1,\dots,X_n$). Then it suffices to consider these last (finite number) of the $X_i$'s with $X_1,\dots,X_n$ as your finite cover.

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Do you need more details? – user17090 Nov 6 '11 at 18:08
What exactly is the argument here? All I can guess is "if $A \subset X$ is compact and $X\setminus A$ is dense in $X$, then $X$ is compact". But that's plainly false. – Chris Eagle Nov 6 '11 at 18:20
@Chris: I corrected the argument. – user17090 Nov 6 '11 at 18:54
Note that the two circles is still the Aleksandroff double circle, and the unit closed disk is still compact, so it's a union of compact sets, hence compact. – Henno Brandsma Nov 6 '11 at 18:58

If I understand you right the final space is a finite union of compact sets. Such a union is always compact.

If you have an open cover of the entire space, take a finite subcover of the closed unit disk together with a finite subcover of the double circle. The union of the subcovers is still finite, and must cover everything.

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This is not a (disjoint) union of compact sets since $\{z\colon |z|<1\}$ is not compact. I don't think about this space as a union anymore since it is confusing in this context. I think I have clarified it in my edit. – Daria Morys Nov 6 '11 at 16:34
Who says it has to be disjoint? Unless I'm parsing you wrong it is a (non-disjoint) union of the closed disk and the double circle, each of which are compact. – Henning Makholm Nov 6 '11 at 16:38

I think what the OP means is the following: let $D$ be the closed unit disk in $\mathbb{R}^2$, and let $C_1$ be its boundary, and $C_2$ the circle of radius 2, and $X$ = $D \cup C_2$, where points in the open unit disk $U$ have their usual neighborhoods, a neighborhood of $z$ in $C_1$ is of the form $O \cup p[( O \cap C_1 ) \setminus {z}]$, where $O$ is an open neighborhood of $z$ in the Euclidean subspace topology of $D$, and $p : C_1 \rightarrow C_2$ is the radial projection, and points in $C_2$ are isolated.

This set is similar to the Aleksandroff double circle (as $C_1 \cup C_2$ is just that space) and $U$ has the same topology as it has in the plane, while the spaces are now glued together naturally.

The same argument that shows that the double circle is compact will work here as well. Also it is the union of $D$ and $C_1 \cup C_2$ (non-disjoint, but that does not matter) and both of these are compact (note that $D$ has the same subspace topology as it has in the plane).

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