If $A$ has strictly positive reach, does the set $\{ x \in A \colon B(x,\epsilon) \subseteq A \}$ also?

Let $A \subseteq \mathbb{R}^n$ with $\text{reach}(A) > 0$ (see https://en.wikipedia.org/wiki/Reach_(mathematics) ).

Define for any $\epsilon>0$, the "removal of $\epsilon$-thick boundary",

$$A^{-\epsilon} = \{ x \in A \colon B(x,\epsilon) \subseteq A \}.$$The question is: Does $A^{-\epsilon}$ again have strictly positive reach?

Note that if one replaces 'strictly positive reach' with 'convex', then the assertion is true.

I know that in general strictly positive reach is not closed under intersections.However, I have tried to come up with counterexamples but with no luck. This has then led me to believe the answer might be yes since maybe one can use the point in $A^{-\epsilon}$which is closest to $\partial A$. However, then this would mean that the reach would increase for $A^{-\epsilon}$, which I am a bit skeptical of...

Asking ChatGPT4 didn't provide additional insight either.