Remove the closure of simply connected region from the interior of a simply connected region. Is it true that the resulting domain can be mapped conformally to some annulus?
$\begingroup$
$\endgroup$
1
-
1$\begingroup$ Another "counter-example", take whole C^1 and remove a disk from it... $\endgroup$– Dmitri PanovJan 8, 2010 at 23:46
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
The answer is yes. This is a special case of theorem 10 in Ahlfors' Complex Analysis, section 5, chapter 6. (Special in that the theorem more generally says that if the complement of the domain has $n$ connected components not reduced to points in the extended plane, then the domain is equivalent to an annulus from which $n-2$ concentric slits have been removed. In your case $n=2$.)
answered Jan 8, 2010 at 20:47
$\begingroup$
$\endgroup$
0
No. The resulting set need not even be connected. And if it is, it need not be doubly connected, as the interior region may have boundary points in common with the original region. Aside from these crude objections, however, the answers of Mariano and Scott are OK.
answered Jan 8, 2010 at 20:56
$\begingroup$
$\endgroup$
1
See Wikipedia. The third entry of the list gives an affirmative answer.
answered Jan 8, 2010 at 20:48
-
2$\begingroup$ Thanks for the lesson, Mariano and Scott. Henceforth I will look in standard references before asking. $\endgroup$– AnweshiJan 8, 2010 at 20:51