If $A,B$ are compacts,, convex with nonempty interiors, prove that they are homeomorphic
My ideia is to find a homeomorphism between $A$ and $B$ and extended to closure, since in this case $\text{int}(\overline{A}) = \text{int}(A)$ (Can I do this?). My question is: prove that $\text{int}(A)$ and $\text{int}(B)$ are homeomorphic? I'm trying to prove that a open convex set in $\mathbb{R}^{n}$ is homeomorphic to an open ball, but I still cannot. Can someone give me a hint?
I didnt look for other questions because I dont want to see the complete solution. I just wanted a hint of how to prove it.
Lastly, is possible to generalize this result for a complete metric space?
Hint: Without lost of generality assume $0 \in int A $. Then consider $f: A \to B$ with
$$ f(x)= \frac{x }{\| x \|} \quad \forall x \in bd A $$
and $f(0) = 0 .$ For the case $x \in int A \setminus \{0\}$, note that $f^{-1} \{\frac{x }{\| x \|} \} \in bd A $ is singleton so lets call it by $\eta (x)$ then define $$ f(x) = \frac{x}{\| \eta (x) \|} $$
Here $B$ is the unit closed ball.
It is easy to show that $f$ is bijection. Continuity of $f$ follows from continuity of $\eta : bd B \to bd A$. And no need to prove the continuity of the inverse of $f$ because $A$ is compact.