We say that a compact subset $E$ of the Riemann sphere $\mathbb{C}_\infty$ is (conformally) removable if every homeomorphism of $\mathbb{C}_\infty$ conformal outside $E$ is actually conformal everywhere, i.e. is a Mobius transformation.

My question is the following :

Suppose $\Gamma$ is a non-removable Jordan curve in the plane. Does $\Gamma$ necessarily contains a non-removable proper closed subset?

Thank you, Malik

EDIT As mentioned by Igor Rivin, in the paper Some homeomorphisms of the sphere conformal off a curve, Chris Bishop raises the above question. However, the paper is 20 years old and I talked to Chris about it, and even back then he really wasn't sure that the problem was open. The question simply did not seem trivial to him.

EDIT 2 It is a direct consequence of the measurable Riemann mapping theorem that every set of positive area is non-removable, so the answer to the question is yes if $\Gamma$ has positive area. What if the area of $\Gamma$ is zero?


1 Answer 1


In 1994 this was open, as evidenced by this paper by Chris Bishop (which has a nice survey). (Some homeomorphisms of the sphere conformal off a curve).

  • 4
    $\begingroup$ Ah yes, I should have mentioned it, but I asked Chris about it, and he told me that he raised this question 20 years ago, because it didn't seem trivial to him, but he really wasn't sure if it was open or not. Since the question is so natural, I expect that someone knows something about it, especially considering the recent interest in removable curves. $\endgroup$ Sep 16, 2014 at 21:12
  • $\begingroup$ Ah, someone downvoted this. No one ever went broke underestimating human nature... $\endgroup$
    – Igor Rivin
    Sep 26, 2014 at 16:31

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