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 $z\mapsto \frac{1}{\phi(z)}$ frac{1}{\phi(1/z)}$then is a homeomorphism between the earring defined with$a_i$and the standard one. Post Undeleted by Zavosh 3 typo Consider$\mathbb{R}^2 \subset \mathbb{C}\cup\infty$. mathbb{C}\cup i \infty$. 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 $2n+i\mathbb{R}$ \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(z)}$then is a homeomorphism between the earring defined with$a_i\$ and the standard one.