Are all Hawaiian Earrings homeomorphic?

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The Hawaiian Earring is usually constructed as the union of circles of radius 1/n centered at (0,1/n): $\bigcup_1^\infty \left[ (0, \frac{1}{n}) + \frac{1}{n}S^1 \right]$. However, nothing stops us from using the sequence of radii $1/n^2$ or any other sequence of numbers $a_n$.

I will call a Hawaiian Earring for a sequence (of distinct real numbers), A = {an}, the union of circles of radius an centered at (0, an). Let the union inherit its topological structure from R2. Are all of these spaces homeomorphic?

If an is a monotone decreasing sequence converging to 0, is its Hawaiian Earring homeomorphic to that of the sequence {1/n}?

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This is a comment about the tag. Nobody studies Hawaiian earring in general topology. The first encounter is when you study algebraic topology -- more precisely, the fundamental group. – Anweshi Jan 6 2010 at 0:53

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Right, but it is not necessary to study algebraic topology to determine the (general) topological properties of the Hawaiian earring and/or to use it as a counterexample in (general) topology. – Qiaochu Yuan Jan 6 2010 at 2:57
What counterexample in general topology did you use the Hawaiian ear ring for? I would like to know. – Anweshi Jan 6 2010 at 12:50

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I think Anweshi's right in that the Hawaiian earring is not at all pathological from the viewpoint of general topology (hence not a probable source of counterexamples): it's a compact, connected, locally path-connected subset of the Euclidean plane. I make this point in the commentary in John Armstrong's blog (linked to in my answer below). However, this particular question is a question about general topology, right? – Pete L. Clark Jan 6 2010 at 12:55

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@Pete and Anweshi, I origionally came across the Hawaiian earring in generaly topology. It was given as an example of somthing which is compact, connected, locally path-connected but NOT locally simply connected (due to problems at (0,0)) i.e. as a counter example to the idea that locally path-connected implies locally simply connected. – qwerty1793 Aug 10 2010 at 22:17
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4 Answers

The Hawaian earring is the one-point compactification of a countable union of open intervals. This description is independent of the radii used to construct it.

A beautiful reference about this space is [Cannon, J. W.; Conner, G. R. The combinatorial structure of the Hawaiian earring group. Topology Appl. 106 (2000), no. 3, 225--271 MR1775709) If I recall correctly, they prove my claim there.

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See Example 1.25 in "Algebraic Topology" by Allen Hatcher. The Hawaiian Earring has a different fundamental group than the infinite wedge of circles. – John Mangual Jan 5 2010 at 20:57

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In the wedge of infinitely many circles, you can construct a sequence (whose points are the antipodal points in each circle, of the joining point) which does not have a convergent subsequence: the wedge is, therefore, not compact. – Mariano Suárez-Alvarez Jan 5 2010 at 20:57
That is the wedge of infinitely many circles. Isn't that non-homeomorphic to the Hawaiian? – Anweshi Jan 5 2010 at 20:57
Oops, I screwed by deleting my comment and re-posting in a different form.. The one of mine above, should have been the first. Sorry. – Anweshi Jan 5 2010 at 20:59

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@Anweshi: the special point in an infinite wedge of circles does not have a countable basis of neighborhoods, so in particular that wedge is not metrizable and, as a consequence, it is not a subspace of $\mathbb R^2$. – Mariano Suárez-Alvarez Jan 5 2010 at 21:03
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Mariano is absolutely right. [And so, for the record, is Joel.]

Coincidentally, an equivalent question came up on the blogosphere in late 2008, and I answered it. The original website is

http://mathphdthoughts.blogspot.com/2008/09/hawaiian-earring.html

Here is my post on that page:

In order to understand the difference [between the Hawaiian earring and the wedge of circles, that is -- 1/5/10] you need to look explicitly at the definition of the topology on a CW-complex with infinitely many cells. By definition, this is the direct limit over the topologies of the subcomplexes with only finitely many cells.

In other words, a subset of the CW complex is open iff its intersection with each individual cell is open. (Another way of saying this is that this is the strongest topology on the entire complex such that each of the inclusion maps from the cells is continuous. This is an instance of a "final topology." Why it is also sometimes called a "weak topology" is not so clear to me: the only reasonable explanation is that the meanings of 'weak' and 'strong' used to be the reverse of what they now are, which I believe is unfortunately the case.)

Anyway, to compare the Hawaiian earring to the infinite bouquet of circles, look at the neighborhood bases of the central point P. On the Hawaiian earring, any open set containing P must contain the entire nth circle for all sufficiently large n, and for the remaining finitely many circles must contain an open interval about P on that circle. However, on the bouquet of circles, the neighborhoods of P are exactly the subsets which contain an open interval around P on each circle. This is a much larger collection of neighborhoods, and indeed the CW-topology is strictly finer than the earring topology.

From this it is easy to see that the CW-topology is not compact, in any number of ways:

(i) Find a closed, discrete infinite subset.

(ii) Note that it is Hausdorff and apply the fact (which can be found in Rudin's Real and Complex Analysis) that any two compact Hausdorff topologies on the same set are incomparable.

(iii) Convince yourself that any CW-complex is compact iff it has finitely many cells.

This generated some discussion on John Armstrong's blog:

http://unapologetic.wordpress.com/2008/09/12/hawai%ca%bbian-earrings/#comments

The very last comment mentions that the earring as the one-point compactification of a countably infinite disjoint union of open intervals. This is a nice observation, the more so since it's completely obvious: the earring is a closed, bounded subset of the Euclidean plane, hence compact. Remove the central point from it and you do indeed get an infinite disjoint union of open intervals. (And, of course, the one-point compactification of a locally compact Hausdorff space is unique up to unique isomorphism.)

This makes clear that the homeomorphism type of the earring does not depend on the radii of the circles -- so long as they converge to 0, of course. (Note that the monotonicity is a superfluous hypothesis. If a sequence converges to $0$, you can reorder it so as to be monotonically decreasing, and the resulting subset of the plane can't tell the difference.)

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You've been caught red handed!  Locally "compact Hausdorff" indeed!  – Harry Gindi Aug 20 2010 at 4:36
Well, if red-handed counts for things that happened over seven months ago... – Harry Gindi Aug 20 2010 at 4:37

The answer to your first question is No, they are not all homeomorphic. In the first question you did not insist that the an converge to 0, and so let us entertain the idea of other crazy sequences. For example, we might let an enumerate all the rational numbers. In this case, we would have circles of every rational radius. This is clearly not homeomorphic to the ordinary Hawaiian earring. For example, every convergent sequence in the ordinary Hawaiian earing lays on a path, but this is not true for the crazy dense version, since every point will be a limit of points on other circles.

You can make a less-crazy counterexample by having just two limit points in the sequence an. For example, let a2n converge to 1/2 and a2n+1 converge to 0. This example would be compact, but still different from the classical earring.

A similar arguent shows that any two sequences with different finite numbers of limit points will be non-homeomorphic. I believe that the homeomorphism type of the resulting earring will be determined by the homeomorphism type of the set {an}, plus the question of whether 0 is a limit point.

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Well, that one isn't compact, so it isn't the same as the converging to 0 case, right? But you're right, this info will affect the homemorphism type. – Joel David Hamkins Jan 6 2010 at 0:29
Wait a minute. I think the case of converging to infinity is the same as having any nonconvergent sequence, no? I don't think infinity is special in the way that 0 is special in this construction. – Joel David Hamkins Jan 6 2010 at 0:45
But the homeomorphism type of a sequence converging to infinity is the same as the homeomorphism type of a sequence converging to a finite number (but not including that number). – Pete L. Clark Jan 6 2010 at 0:50
Wait -- no, you're right again.  The sequences $a_n = 2 - \frac{1}{n}$ and $a_n = n$ give rise to homeomorphic earrings.   – Pete L. Clark Jan 6 2010 at 0:53
Yes, I think that's right. – Joel David Hamkins Jan 6 2010 at 1:13
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Consider $\mathbb{R}^2 \subset \mathbb{C}\cup i \infty$. The inversion $z \mapsto \frac{1}{z}$ sends the circles $(a_n,0)+a_nS^1$ to the vertical lines $\{\frac{1}{2a_n}+i\mathbb{R}\}\cup i\infty$. Let $b_i=\frac{1}{a_i}$ for $i\geq 1$ and $b_0=0$. The function $f(x)=b_{\lfloor x \rfloor} + \{x\}(b_{\lfloor x + 1\rfloor} - b_{\lfloor x \rfloor})$ is a homeomorphism of $\mathbb{R}^+$ that sends the positive integers to $b_1, b_2,...$. Letting $\phi(x+iy)=\frac{1}{2}f(2x)+iy$, one gets an automorphism of the right-hand plane that sends the vertical lines $\frac{n}{2}+i\mathbb{R}$ to $\frac{1}{2a_n}+i\mathbb{R}$. Extending $\phi$ so that it sends $i\infty$ to $i\infty$, the function $z\mapsto \frac{1}{\phi(1/z)}$ then is a homeomorphism between the earring defined with $a_i$ and the standard one.

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