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Zavosh
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Consider $\mathbb{R}^2 \subset \mathbb{C}\cup\infty$. 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 $\phi(x+iy)=f(x)+iy$, one gets an automorphism of the right-hand plane that sends the vertical lines $\frac{1}{2a_n}+i\mathbb{R}$ to $\frac{1}{2n}+i\mathbb{R}$. Extending $\phi$ so that it sends $i\infty$ to $i\infty$, the function $z\mapsto \frac{1}{\phi(z)}$ then is a homeomorphism between the earring defined with $a_i$ and the standard one.

Zavosh
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