TITLE: Are all Hawaiian Earrings homeomorphic?
QUESTION [31 upvotes]: 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), $A = \lbrace a_n \rbrace$, 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 $\lbrace 1/n \rbrace$?

REPLY [4 votes]: 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.