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Nov
29
comment Classification of open subset of $\mathbb{R}^{3}$
@google You're welcome, I'm glad this helps.
Nov
28
comment Classification of open subset of $\mathbb{R}^{3}$
@AlexandreEremenko Great, thank you for the precision.
Nov
28
comment Classification of open subset of $\mathbb{R}^{3}$
@AlexandreEremenko Who proved the theorem in the general case? I always thought it was Grötzsch, Uber das Parallelschlitztheorem der konformen Abbildung schlichter Bereiche, Berichte Leipzig (84) (1932) 15-36.
Nov
27
comment Classification of open subset of $\mathbb{R}^{3}$
I think you can find a proof of this result in Goluzin's book "Geometric theory of functions of a complex variable". I don't have it with me though so I cannot say for sure.
Nov
27
comment Classification of open subset of $\mathbb{R}^{3}$
The result on uniformization by slit domains is indeed due to Grotzsch, Uber das Parallelschlitztheorem der konformen Abbildung schlichter Bereiche, Berichte Leipzig (84) (1932) 15-36. The theorem says that any domain in the plane is conformally equivalent to a domain whose complementary components are all points or compact horizontal segments. The number of complementary components might be uncountable.
Oct
14
revised A question on Ahlfors covering surface
deleted 177 characters in body
Oct
14
comment A question on Ahlfors covering surface
@AlexandreEremenko Ah yes indeed, I corrected, thanks. I was referring to the proof in the second paper.
Oct
14
answered A question on Ahlfors covering surface
Oct
11
comment Are the algebraic numbers dense everywhere on the boundary of the Mandelbrot set?
These are called Misiurewicz points, right? It might be useful to add this to your answer in order to facilitate literature search.
Oct
5
awarded  Nice Answer
Sep
23
accepted How bad can a circle domain get?
Sep
23
comment How bad can a circle domain get?
Your argument is indeed quite reminiscent of the Baire theorem, but thank you for the details and also for the outline of the construction for Question 2. As usual, your answer is quite valuable!
Sep
23
awarded  Nice Question
Sep
22
comment How bad can a circle domain get?
If yes, then your example of a circle domain $\Omega$ should also settle Question 1, unless I am missing something. Indeed, if $\partial \Omega$ were the union of countably many circles, countably many Cantor sets and countably many points, then one of these sets, say $A$, would have nonempty interior in $\partial \Omega$, by Baire. This means that there would exist some open set $U$ with $U \cap \partial \Omega \neq \emptyset$ and $U \cap \partial \Omega \subset A$. But then any point in $U \cap \partial \Omega$ would be isolated from circles... Am I missing something?
Sep
22
comment How bad can a circle domain get?
Thank you, I'll look into Klein combinations. Just to make sure : in your example, is any point on a circle also a limit point of infinitely many circles?
Sep
22
comment How bad can a circle domain get?
@EricWofsey I do not consider such points to be limits of circles. For me, limits of circles mean infinitely many circles approaching the point.
Sep
22
comment How bad can a circle domain get?
@EricWofsey No, not necessarily.
Sep
22
revised How bad can a circle domain get?
added 334 characters in body
Sep
22
comment How bad can a circle domain get?
In other words : If a circle in the boundary of $X$ is not isolated from point boundary components, then in the quotient space it will not correspond to an isolated point in the countable set.
Sep
22
comment How bad can a circle domain get?
@MathieuBaillif Yes, it is true that the quotient space is homeomorphic to $\widehat{\mathbb{C}}$. It is a very special case of Moore's theorem on upper semi-continuous decompositions of the sphere. However, here I want the set of point components to be a countable union of Cantor sets and points, and one runs into trouble with this argument because the set of point components need not be closed...