Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $W_i\cap \{z\colon |z|=1\}=V_i$.

share|improve this question
"Alexandrov double circle" is not very googlable. Perhaps it would help if you added an appropriate link? It seems that you're assuming that the reader knows some specific topological properties that you imply by using that term. –  Henning Makholm Nov 6 '11 at 16:06
I think the EDIT is not clear enough. And things are more complicated than the answer below suggests. –  user17090 Nov 6 '11 at 16:47
@Ali Bleybel, I hope it's clear enough now. –  Daria Morys Nov 6 '11 at 17:02
First I need an explanantion why the original construction is compact. I mean, the larger circle has discrete topology and of size continuum so, if I am not mistaken, it is not compact. –  user17090 Nov 6 '11 at 17:34

3 Answers 3

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.

share|improve this answer
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.

share|improve this answer
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).

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.