I've been thinking about this statement for a while, and I think it's true, but I'm not sure of how to prove it. The statement is

A Jordan curve $J$ that is symmetric about the origin $p$ does not pass through $p$.

I think one should be able to argue by contradiction that if $J$ passes through the origin, then $J$ has a self-intersection, which contradicts the fact $J$ is a Jordan curve, but I'm having trouble with the details.

  • $\begingroup$ How does one formalize "symmetric about the origin" for a Jordan curve? If curve goes through $(x,y)$ then also goes through $(-x,-y)$? $\endgroup$ – coffeemath Apr 22 '19 at 7:48
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    $\begingroup$ @coffeemath "Symmetric about the origin" means the curve is unchanged when reflected across both the x-axis and y-axis, which is exactly what you have described. $\endgroup$ – YuiTo Cheng Apr 22 '19 at 7:50
  • $\begingroup$ It seems like you should be able to show that the interior is symmetric about the origin, so it contains some disk about the origin. $\endgroup$ – saulspatz Apr 22 '19 at 8:28
  • $\begingroup$ Hint: Show first that a homeomorphism if order 2 of the real line has a fixed point. Then argue that Jordan curve minus a point is homeomorphic to ... $\endgroup$ – Moishe Kohan Apr 22 '19 at 11:50
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    $\begingroup$ @MoisheKohan Would you expand on this a bit, please? I don't see it. To begin with, does "order $2$" mean $\phi\circ\phi = id?$ $\endgroup$ – saulspatz Apr 22 '19 at 12:30

OK: Suppose that $J\subset R^n$ is a Jordan curve invariant under the antipodal map $\phi: x\mapsto -x$ and $0\in J$. Then $\phi$ restricts to an involution $\tau$ on $J-\{0\}\cong {\mathbb R}$: $\tau\circ \tau=id$. Since $\phi$ has only one fixed point, $\tau$ has no fixed points. Now, contradiction comes from the following lemma

Lemma. Let $f: {\mathbb R}\to {\mathbb R}$ be a homeomorphism of finite order $k$, i.e. $f^k=id$. Then $f$ has at least one fixed point.

I will leave you a proof as an exercise, compare this question.

  • $\begingroup$ Very slick! This makes sense. Thank you. $\endgroup$ – Zain Siddiqui Apr 23 '19 at 15:42

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