Let us consider metric space $(X, d)$. Define closed $\epsilon$-fattening of a closed (and bounded) set $A$ by: $$ A_\epsilon = \{ x \in X : \exists a \in A \quad d(a, x) \le \epsilon \}. $$
What I want to know is when statement $$ A \subset B_\epsilon \land B \subset A_\epsilon \iff A_\lambda \subset B_{\epsilon + \lambda} \land B_\lambda \subset A_{\epsilon + \lambda}$$
is true for every $\lambda \ge 0$, $A, B \subset X$, by imposing some properties on $(X, d)$.
It isn't hard to prove that implication $\implies$ always holds, regardless of properties of $(X, d)$. So far, I was able to find counterexample to implication $\impliedby$ in:
- bounded space: let $(X, d) = ([0,1], d_e)$ Then $A = \{0\}$ and $B = \{1\}$ gives counterexample.
- 1-connected space: Let us consider space $(X, d)$ where $X$ is a subset of $\mathbb{R}^2$ consisting of every point which would be covered by moving unit circles from point $(0, -2)$ to $(2, 0)$, from point $(0, 2)$ also to $(2, 0)$ and from $(2, 0)$ infinitely to the right. (Created shape would look Y with infinitely elongated bottom line, placed on its side.) $d$ is given by lengths of shortest paths connecting two points. (Paths have to be within $X$). Then $A = B((0,-2), 1)$ and $B = B((0,2), 1)$ give counterexample for large enough $\lambda$ and $\epsilon = 0$.
I hope that $\impliedby$ is true for normed spaces, however so far I was unable to prove it.