(a) Find a metric space in which the boundary of $r$-neighborhood of $p$ is not equal to sphere of radius $r$ at $p$
(b) Need the boundary be contained in the sphere?
My attempt.
(a) I think in a discrete space $M$ with the discrete metric. Thus, if $V_{r}({p})$ is the $r$-neighborhood at $p$, $\partial V_{r}(p) = \overline{V_{r}(p)}-\mathrm{int}V_{r}(p)$. Since every subset $M$ is clopen, $\partial V_{r}(p) = \emptyset$. Take $M = \mathbb{N}$ and $r = 1$ its enough?
(b) I try to write $S_{r}(p) = \overline{B_{r}(p)} - B_{r}(p)$. But $V_{r}(p) \subset B_{r}(p)$ and so, $\overline{V_{r}(p)} \subset \overline{B_{r}(p)}$. Then, $$\overline{V_{r}(p)}-V_{r}(p) \subset \overline{B_{r}(p)} - B_{r}(p)??\tag{$\ast$}$$ I think that I can have a point $p \in B_{r}(p)$ and $p \not\in B_{r}(p)$ and so, this $p \in \overline{V_{r}(p)}-V_{r}(p)$ and $p \not\in \overline{B_{r}(p)} - B_{r}(p)$. But I couldn't find an explicit counterexample.
Can someone help me?
For any open ball $B(p,r)$, its boundary $\partial B(p,r)$ is always contained in the sphere $S(p,r)$.
Indeed, first notice that the closure $\overline{B(p,r)}$ of the open ball is always contained in the closed ball $\overline{B}(p,r)$. This follows from the fact that $\overline{B}(p,r)$ is a closed set.
For any set $A$ holds $\partial A = \overline{A} \setminus \operatorname{Int}(A)$ so we have $$\partial B(p,r) = \overline{B(p,r)} \setminus B(p,r) \subseteq \overline{B}(p,r) \setminus B(p,r)=S(p,r) $$