Prove that all the open convex (non-empty) subset of $\mathbb{R}^n$ are homeomorphic to each other.
First of all I'm not sure if the statements is correct in the first place or not. I sketched a proof, with the additional hypothesis that the sets are bounded, that I'm unable to complete. I would like to complete the proof (in the bounded case) and to generalize it to the unbounded case.
The main idea of the proof is to show that every open convex set is homeomorphic to $B(0,1)$ (A bit of notation : $B(x_0,\rho) := \{ x \in \mathbb{R}^n \; : \; ||x - x_0|| < \rho \} \;,\; B := B(\underline{0},1) \; , \; ||x||$ is the euclidean norm of the vector $x$)
Let $A \subset \mathbb{R}^n$ be a open convex bounded set. WLOG suppose that $\underline{0} \in A$.
Let $r : \mathbb{R}^n \setminus \{\underline{0}\} \to (0,+\infty)$ defined in this way
$r(x) := \sup\{ c > 0 \; : \; c \frac{x}{||x||} \in A \} $
clearly because $A$ is bounded $r(x) < +\infty$
Now let $f : A \to B$ defined in this way
$\begin{cases} f(x) &= \underline{0} \;\; \hspace{0.7cm}(if \;x = \underline{0}) \\ f(x) &= \frac{x}{r(x)}\;\; \hspace{0.2cm}(if \;x \neq \underline{0}) \end{cases}$
I want to show that $f$ is an homeomorphism of $A$ into $B$. I have to prove that $f$ is : bijective, continuos and has continuos inverse.
I've been able to show that $f$ is bijective, and I thought how to proof that $f$ has continuos inverse, this is how : basically I define another function $g : \overline{A} \to \overline{B}$ wich is a continuos extension of $f$. Now because $\overline{A}$ is compact and $\overline{B}$ is Haussdorf, if $g$ is continuos then it's also closed, thus $g^{-1}$ is continuos, thus $f^{-1}$ is continuos, therefore I only have to show that $f$ is continuos (wich would imply that $g$ is continuos).
To do that it's sufficient to show that $f$ is continuos in any point. First I prove that $f$ is continuos in $\underline{0}$. Let $\delta > 0$ such that $B(\underline{0},\delta) \subseteq A$ ($\delta$ exists because $\underline{0} \in A$ and $A$ is open), then by definition $r(x) \geq \delta \;\; \forall x \in \mathbb{R}^n \setminus \{ \underline{0} \}$, therefore $|f(x)| = |\frac{x}{r(x)}| \leq \frac{|x|}{\delta}$, which clearly shows that $\lim\limits_{x \to \underline{0}}{f(x) = \underline{0} = f(\underline{0})}$ thus $f$ is continuos in $\underline{0}$. To show that $f$ is continuos in $A \setminus \{\underline{0}\}$ is enough to show that $r$ is continuos.
I've been only able to show that $r$ is lower semicontinuos, this is how I did that. Let $x_0 \in \mathbb{R}^n \setminus \{ \underline{0} \}$, WLOG let let $\epsilon > 0$, let $y := \frac{x_0}{||x_0||}(r(x_0) - \epsilon)$. Basically $y \in A$ so I can find a proper neighborhood $V$ of $y$ that's contained in $A$ and such that, by "projecting" in some sense this neighborhood $V$ on $x_0$ I get a neighborhood $U$ of $x_0$ such that $x \in U \implies r(x) > r(x_0) - \epsilon \;\;$ (this part is really non rigorous but can be made rigorous, I wrote it like this to make the proof easier)
My problems are : the upper semicontinuity of $r$, which I've been unable to prove, and how to generalize this proof when $A$ is unbounded.
About the unbounded case I thinked about defining $r$ in the same way but with codomain $(0,+\infty]$, in this case I have to define $f$ in a different way, I thinked about something like this (for $x \neq \underline{0}$ )
$f(x) = \frac{x}{||x||}G(||x||,r(x))$
where $G : (0,+\infty) \times (0,+\infty] \to (0,1)$ is a continuos function such that, $\forall \lambda > 0$ $g(t) := G(t,\lambda)$ is a growing homeomorphism of $(0,\lambda)$ into $(0,1)$
(and such that $\lim\limits_{x \to \underline{0}}{f(x)} = \underline{0}$ )
My questions are : is the statemente that I want to show (i.e all open convex sets of $\mathbb{R}^n$ are homeomorphic) true? Is the proof that I wrote correct? How can I show that $r$ is upper semicontinuos?
I know that the proof isn't completely rigorous, in fact I even haven't showed the all proof but all the missing pieces are pretty easy to show so it shouldn't be a big of a problem.
