# A Jordan Curve Symmetric About the Origin Does Not Pass Through Origin

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.

• How does one formalize "symmetric about the origin" for a Jordan curve? If curve goes through $(x,y)$ then also goes through $(-x,-y)$? – coffeemath Apr 22 '19 at 7:48
• @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. – YuiTo Cheng Apr 22 '19 at 7:50
• 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. – saulspatz Apr 22 '19 at 8:28
• 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 ... – Moishe Kohan Apr 22 '19 at 11:50
• @MoisheKohan Would you expand on this a bit, please? I don't see it. To begin with, does "order $2$" mean $\phi\circ\phi = id?$ – 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.