# On interiors of Jordan curves

Recently,while studying a problem in physics whose solution required an application of the Jordan curve theorem to the phase space in order to make it rigorous, I asked myself if its possible for two Jordan curves $$E$$ and $$F$$ on the plane to be contained in each other's interiors.I tried to prove that it's not,failed,realized there might be some counterexample,and ended up drawing complicated spirals all over the pages of my notebook.I haven't yet got it, so I ask for some help here.Properly phrased,my question is,for two Jordan curves $$E$$ and $$F$$ on the plane,it's known that $$E$$ is contained in the interior of $$F$$.Is it possible for $$F$$ to be contained in the interior of $$E$$?

• It's funny, I'm absolutely certain this can't happen but I have no idea how to prove it. – Cronus Jul 9 at 12:12

$$F$$ (resp. $$E$$) separates the plane in two connected components $$U_F$$ and $$B_F$$ (resp. $$U_E$$ and $$B_E$$), where $$U$$ stands for unbounded and $$B$$ for bounded. This is part of the statement of the Jordan curve theorem. The bounded one is what you refer to as "interior". So we know $$E \subset B_F$$.
Assume $$F \subset B_E$$. This implies that $$U_F$$ (and $$B_F$$ too, but this is irrelevant to the argument that follows) intersects $$B_E$$. Since $$E$$ does not intersect $$U_F$$ (as it is contained in $$B_F$$), it follows that $$U_F$$ is contained in the separation $$U_E$$ and $$B_E$$. But we know that $$U_F$$ intersects $$B_E$$ and therefore, since $$U_F$$ is connected, it must be contained in $$B_E$$. But this is a contradiction, due to the fact $$U_F$$ is unbounded and $$B_E$$ is bounded.