MHB Analytic Isomorphism: Annulus vs. Punctured Unit Disc

[Dustinsfl]

Why can't an annulus be analytically isomorphic to the punctured unit disc?

$A_{r,R}$ is an annulus

Theorem: $A_{r,R}$ is analytically isomorphic to $A_{s,S}$ iff $R/r = S/s$.

If our annulus $A_{1,2}$, then $R/r = 2$ and the punctured disc would be $\lim\limits_{s\to 0}1/s = \infty$.

So here is a counter example.

 

[Dustinsfl]

Re: analytic ismoprhism

Why can't an annulus be analytically isomorphic to the punctured unit disc?

$A_{r,R}$ is an annulus

Theorem: $A_{r,R}$ is analytically isomorphic to $A_{s,S}$ iff $R/r = S/s$.

If our annulus $A_{1,2}$, then $R/r = 2$ and the punctured disc would be $\lim\limits_{s\to 0}1/s = \infty$.

So here is a counter example.

How can I do this without invoking this theorem?

 

[Dustinsfl]

Re: analytic ismoprhism

How about this?

Let $f:\mathbb{D}\to A_{1,2}$.
Let $U$ be an open neighborhood of 0 of radius $\epsilon > 0$.
Then $f(U)\subset A_{1,2}$.
Letting $\epsilon$ be small enough we would have that $f(U)$ is not in the annulus.