Timeline for Hausdorff distance on algebraic curves

8 events

when toggle format	what		by	license	comment
Apr 3, 2012 at 15:02	answer	added	anonymous		timeline score: 1
Apr 3, 2012 at 12:39	history	edited	Pablo Shmerkin	CC BY-SA 3.0	Added some new relevant information, clarified some definitions; added 35 characters in body
Mar 31, 2012 at 0:55	comment	added	Pablo Shmerkin		Will - I think you're correct, I've edited the question accordingly.
Mar 31, 2012 at 0:54	history	edited	Pablo Shmerkin	CC BY-SA 3.0	added 126 characters in body
Mar 30, 2012 at 21:15	comment	added	Will Sawin		Why are there discontinuities on reducible curves? My intuition says that there are not. For instance, $xy+ax+by+c=(x+b)(y+a)+c-ab$, so the asymptotic and the distance from them both vary continuously. I agree that there is non-differentiability but that is much less bad.
Mar 30, 2012 at 15:02	comment	added	Pablo Shmerkin		Gerald - no, elements of $A_1$ are (single) straight lines ($A_n$ is the family of zero sets $P=0$ where $P\in\mathbb{R}[x,y]$ has degree at most $n$). For $n=1,2$ the answer is easily be seen to be yes (and the actual dimension can be computed) because linear/quadratic algebraic curves admit a simple explicit classification.
Mar 30, 2012 at 14:38	comment	added	Gerald Edgar		So, for example, an element of $A_1$ could be the union of a finite number of straight lines through the origin? So it would be enough to show that the set of these is infinite-dimensional, when $X$ is the unit disk?
Mar 30, 2012 at 13:49	history	asked	Pablo Shmerkin	CC BY-SA 3.0