I'm trying to prove that the boundary $\partial \Omega$ of any non-empty bounded convex planar set $\Omega$ is homeomorphic to $S^1$.
My current ideas are as follows:
- W.L.O.G assume that $0$ is in the interior of $\Omega$. Any half-line emanating from $0$ intersects $\partial \Omega$ at precisely one point. I have shown that the projection $\displaystyle \pi : \partial \Omega \to S^{1} : x\mapsto \frac{x}{\lVert x\rVert}$ is well-defined, continuous and bijective.
- My issue arises with the inverse map $\displaystyle \pi^{-1}: S^1 \to \partial\Omega:x'\mapsto f(x')x'$ where $f(x) = \sup\lbrace t> 0 \mid tx'\in\Omega\rbrace$. $\pi^{-1}$ is continuous provided that $f:S^{1} \to \mathbb{R}_{>0}$ is continuous.
Note that here I am setting $S^1 = \lbrace (x,y) \mid x^{2}+y^{2} = 1\rbrace$ and am endowing it with the subspace topology from $\mathbb{R}^{2}$.
My approach thus far has been to suppose that $f$ isn't continuous and then showing that $\Omega$ is not convex. And following an argument like:
If $f$ is not continuous at $x$ then $\exists \delta >0$ such that for all $n\in \mathbb{N}$ there exists $y_{n}\in B(x;\frac{1}{n})$ such that $|f(y_{n})-f(x)| \geq \delta$. Then we have that the line $[f(y_{n}),f(y_{n+1})]$ is contained in $\overline{\Omega}$. This implies that...
Am I on to the right thing? I can't seem to reach the contradiction I want.
Remark. It may be a rather silly way of doing it but showing that this $f$ is continuous would greatly help me with some other results so I do hope this is the case!