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?