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.
 A: Take $y\in\partial B(x_0,r)$. Can you have $d(y,x_0)\neq r$? If so, there are two possibilities:


*

*$d(y,x_0)<r$. But then $B\bigl(y,r-d(y,x_0)\bigr)\subset B(x_0,r)$ and so $y\in\operatorname{int}\bigl(B(x_0,r)\bigr)$, which implies that $y\notin\partial B(x_0,r)$.

*$d(y,x_0)>r$- But then $B\bigl(y,d(y,x_0)-r\bigr)\cap B(x_0,r)=\emptyset$. So, $y\notin\overline{B(x_0,r)}$, which implies again that $y\notin\partial B(x_0,r)$.

A: This inclusion is true for $B$ the open ball of radius $r$ centered at $x_0$, for the following reasons : 
1) $\overline{B} \subset \{x\in X\mid d(x,x_0)\leq r\}$ : this is because this set is closed. 
2) If $\mathrm{Int}(B) = B = \{x\in X\mid d(x,x_0)<r\}$. 
So if $x\in \partial B$, $x\in \overline{B}$ so $d(x,x_0)\leq r$, but $x\notin B$ so $d(x,x_0)\geq r$; so $d(x,x_0)=r$. 
A: 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.
