The [Hawaiian Earring](http://en.wikipedia.org/wiki/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), $A = \lbrace a\_n \rbrace$, the union of circles of radius $a\_n$ centered at $(0, a\_n )$. Let the union inherit its topological structure from $\mathbb{R}^2$. Are all of these spaces homeomorphic?


If $a\_n$ is a monotone decreasing sequence converging to $0$, is its Hawaiian Earring homeomorphic to that of the sequence $\lbrace 1/n \rbrace$?



Are all Hawaiian Earrings homeomorphic?
The Hawaian earring is the one-point compactification of a countable union of open intervals (with the coproduct or disjoint sum topology). 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](http://www.ams.org/mathscinet-getitem?mr=MR1775709)) If I recall correctly, they prove my claim there.


