Does $\partial B(x_0, r) \subseteq \{x \in X : d(x_0,x) = r \}$ hold in an arbitrary metric space? Let $X$ be a metric space, and let $B(x_0, r)$ denote the open ball of radius $r$ centred at $x_0 \in X$.
Does the statement $\partial B(x_0, r) \subseteq \{x \in X : d(x_0,x) = r \}$ hold true always?
A similar question posted in Math StackExchange
As answered in the link above, we know that the boundary is not the set $\{x \in X : d(x_0,x) = r \}$. however, it seems that the statement $\partial B \subseteq \{x \in X : d(x_0,x) = r \}$ is true. Considering the example in the link above, the boundary of any set in the discrete metric is the empty set which is a subset of $\{x \in X : d(x_0,x) = r \}$. But I am not too certain. Anyone mind showing me some counterexample? Thank you.

Let $(X,d)$ be a metric space, and let $B(x,r)\subseteq X$ be an open ball. Since $\text{Int}\left(B(x,r)\right)=B(x,r)$ (an open ball is an open set) and $\text{Cl}\left(B(x,r)\right)\subseteq\{y \in X :d(y,x)\leq r\}$ (a closed ball is a closed set), it follows that we have 
\begin{aligned}\partial B(x,r)=\text{Cl}\left(B(x,r)\right) \setminus B(x,r) &\subseteq \{y \in X :d(y,x)\leq r\} \setminus B(x,r) \\&=\{y \in X :d(y,x)= r\}.\end{aligned}
Therefore, $\partial B(x,r) \subseteq \{y \in X :d(y,x) = r\}$ holds for any metric space.
