Alexandrov double circle

Question

Let $K$ be the Alexandrov double circle determined by the two circles on the complex plane, $\{z: |z|=1\}$ and $\{z: |z|=2\}$ respectively.

Consider the union of K with the disc $\{z: |z|<1 \}$ with the usual Euclidean topology. Is this space compact? I believe it is, since the topology of the smaller circle is inherited from the complex plane.

EDIT: My "construction" above is a bit vague. Here is a detailed version of what I want:

I want to redefine the Alexandrov double circle starting with the closed unit disc (and project its boundary onto the bigger circle) instead of the unit circle itself. I am not sure if compactness of the resulting space is being preserved.

EDIT2: The open neighbourhoods for points from $\{z\colon |z|<1\}$ are inherited from $\{z\colon |z|\leq 1\}$ (with relative Euclidean topology). Open neighbourhoods for a point $w\in\{z\colon |z|=1\}$ are of the form $U_i = W_i \cup p(V_i\setminus \{w\})$ where $V_i$ and $p$ are the same as in Engelking's book and $W_i$ is any open set set in the closed unit circle such that $W_i\cap \{z\colon |z|=1\}=V_i$.

Answer 1

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.

Answer 2

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.

Answer 3

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).