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Oct
14 |
revised |
A question on Ahlfors covering surface
deleted 177 characters in body |
Oct
14 |
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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 |
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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 |
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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... |
Sep
22 |
awarded | Custodian |
Sep
22 |
revised |
How bad can a circle domain get?
deleted 1 character in body |
Sep
22 |
reviewed | Approve How bad can a circle domain get? |
Sep
22 |
asked | How bad can a circle domain get? |
Aug
21 |
comment |
Absolute continuity and the Luzin N-Property for functions of two variables
Hi Trevor. I don't know about analogs of absolute continuity in several variables, but a very interesting family of continuous functions in $\mathbb{R}^n$ which preserve sets of zero $n$-dimensional Lebesgue measure are the so-called quasiregular mappings. A standard reference for this is the book of Rickman, Quasiregular mappings. You could have a look and figure out precisely what properties are needed in order to preserve zero-measure sets. |