Consider the Euclidean ball $B^n(x,r)$ in $\mathbb{R}^n$ given by:
$$B^n(x,r) = \{z\in\mathbb{R}^n : ||z-x||_2 \leq r\}$$ with centre $x\in\mathbb{R}^n$ and radius $r\geq 0$.
Now we break this ball by removing an arbitrary subset from the boundary of $B^n(x,r)$. Is the resulting set still convex?
My intuition says Yes, and it's easy to visualise this in $\mathbb{R}^2$ or $\mathbb{R}^3$. I'm wondering how I can prove it generally, i.e. for $\mathbb{R}^n$ by considering an arbitrary subset $S\subseteq B^n(x,r)$? If $S = B^n(x,r)$ then the resulting set is empty, which is convex – so we can ignore that situation. We assume $S\subset B^n(x,r)$. What's next?
I consider two points $y_1,y_2 \in B^n(x,r)\backslash S$, and want to show that for $t\in [0,1]$ we have $ty_1 + (1-t)y_2 \in B^n(x,r)\backslash S$. How do I take it from here? I'm stuck.
Thanks!
Addendum:
What if we consider the $p$-norm in general, instead of the Euclidean norm? How do things change with $p$ if the ball is defined as follows?
$$B^n(x,r) = \{z\in\mathbb{R}^n : ||z-x||_p \leq r\}$$
It sounds fun breaking balls in higher dimensions under different norms!
Best Answer
To simplify notation, I'll assume, without loss of generality, that the center of the ball is at the origin. Suppose, toward a contradiction, that you have a counterexample in some high dimension. Say your set contains $p$ and $q$ but not some point $tp+(1-t)q$ on the line segment joining $p$ to $q$. Then this is also a counterexample in the $2$-dimensional subspace spanned by (the vectors from the origin to) $p$ and $q$. So, once you know the result for $2$ dimensions, it follows for all higher dimensions.