Malik Younsi
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 1h answered A question on Ahlfors covering surface 2d 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... 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. Aug 19 comment Is there a reference for “computing $\pi$” using external rays of the Mandelbrot set? This doesn't answer your very interesting question (and surely you are already aware of what follows), but a good reference for this $\pi$ phenomenon is the paper " $\pi$ in the Mandelbrot set" by Klebanoff. In the conclusion, Klebanoff conjectures that there are infinitely many routes at each of the infinitely many pinches of the Mandelbrot set which lead to $\pi$ in this way. Aug 13 awarded Yearling