I am currently trying to understand the proof of Proposition 2 in https://arxiv.org/abs/1904.01681 (page 14).
In there one tries to prove that the sets $A:=\overline{B}_r(0)$ and $B:=\overline{B}_t(0)\setminus B_s(0)$ in $\mathbb{R}^d$ for $0<r<s<t$, are not linearly separable, even after applying a homeomorphism $\phi$.
Heuristically I understand this, as a homeomorphism can never "tear the outer ring apart". But I am having trouble understanding a specific part of the proof. Namely the one where it says that if $\phi(B_s(0))$ and $\phi(\partial B_s(0))$ are not linearly separable, then neither are $\phi(A)$ and $\phi(B)$. Since $A\subset B_s(0)$, intuitively, "shrinking" the set should make linear separability possible right? I can't wrap my head around their argument.
Can someone help out?
You’re right; that argument in the proof is flawed. The reason the proof works is not that $\phi(B_s(0))$ as a whole is not linearly separable from $\phi(\partial B_s(0))$ (as you rightly point out, that could be solved by separating only $\phi(A)$ and not all of $\phi(B_s(0))$) but that no point in $\phi(B_s(0))$ is linearly separable from $\phi(\partial B_s(0))$; since crucially $\partial B_s(0)\subset B$ (and not just $B\subset\partial B_s(0)$), this shows that not just $\phi(B_s(0))$ but any subset of it (and thus $\phi(A)$) is not linearly separable from $\phi(\partial B_s(0))$ and thus from $\phi(B)$).