When is the closure of an open ball equal to the closed ball? It is not necessarily true that the closure of an open ball $B_{r}(x)$ is equal
to the closed ball of the same radius $r$ centered at the same point $x$. For a quick example, take $X$ to be any set and define a metric
$$
d(x,y)=
\begin{cases}
0\qquad&\text{if and only if $x=y$}\\
1&\text{otherwise}
\end{cases}
$$
The open unit ball of radius $1$ around any point $x$ is the singleton set $\{x\}$. Its closure is also the singleton set. However, the closed unit ball of radius $1$ is everything. 
I like this example (even though it is quite artificial) because it can show that this often-assumed falsehood can fail in catastrophic ways. My question is: are there necessary and sufficient conditions that can be placed on the metric space $(X,d)$ which would force the balls to be equal? 
 A: 
More general, we can prove that for any normed space $(V,\|\cdot\|)$, if $x_0\in V$ and $R>0$, then $$\overline{B(x_0;R)}^{\|\cdot\|}=B[x_0;R],$$ where $\overline{B(x_0;R)}^{\|\cdot\|}$ is the closure of the open ball $B(x_0;R)$ in topology induced by $\|\cdot\|$.

The inclusion $\overline{B(x_0;R)}^{\|\cdot\|}\subseteq B[x_0;R]$ is trivial, because $B[x_0;R]$ is a closed set that contains $B(x_0;R)$ and $\overline{B(x_0;R)}^{\|\cdot\|}$ is the smallest closed set that contains $B(x_0;R)$.
To show that $B[x_0;R]\subseteq\overline{B(x_0;R)}^{\|\cdot\|}$, we can prove that any element of $B[x_0;R]$ is the limit of some sequence in $B(x_0;R)$. Let $x\in B[x_0;R]$, then of course $$\|x-x_0\|\leqslant R.$$ Now defines, for each $n\in\Bbb{N}$, $$x_n:=\left(1-\frac{1}{n}\right)x+\frac{1}{n}x_0.$$
Claim 1: The sequence $(x_n)$ converges to $x$ in the norm $\|\cdot\|$. In fact, \begin{eqnarray}
\|x_n-x\| & = & \left\|\left(1-\frac{1}{n}\right)x+\frac{1}{n}x_0-x\right\| \\
& = & \frac{1}{n}\underbrace{\|x-x_0\|}_{\leqslant R} \\
& \leqslant & \frac{R}{n}\longrightarrow0,\quad\text{when}\ n\to\infty,
\end{eqnarray} which proves that $x_n\to x$ in $\|\cdot\|$.
Claim 2: For all $n\in\Bbb{N}$, $x_n\in B(x_0;R)$, that is, $(x_n)$ is a sequence in the set $B(x_0;R)$. This is obviously by the construction of $x_n$, since \begin{eqnarray}
\|x_n-x_0\| & = & \left\|\left(1-\frac{1}{n}\right)x+\frac{1}{n}x_0-x_0\right\| \\
& = & \left\|\left(1-\frac{1}{n}\right)x-x_0\left(1-\frac{1}{n}\right)\right\| \\
& = & \left\|\left(1-\frac{1}{n}\right)(x-x_0)\right\| \\
& = & \left(1-\frac{1}{n}\right)\underbrace{\|x-x_0\|}_{\leqslant R} \\
& \leqslant & \underbrace{\left(1-\frac{1}{n}\right)}_{<1}R<R,
\end{eqnarray} which gives us $x_n\in B(x_0;R)$.
So we conclude that any element of $B[x_0;R]$ is the limit of a sequence in $B(x_0;R)$, hence $B[x_0;R]\subseteq\overline{B(x_0;R)}^{\|\cdot\|}$.
A: Here is a characterization that is straight from the definitions, but which it seems may be useful when verifying that a particular space has the property.
For any metric space $(X,d)$, the following are equivalent:

*

*For any $x\in X$ and radius $r$, the closure of the open ball of radius $r$ around $x$ is the closed ball of radius $r$.

*For any two distinct points $x,y$ in the space and any positive $\epsilon$, there is a point $z$ within $\epsilon$ of $y$, and closer to $x$ than $y$ is.
That is, for every $x\neq y$ and $\epsilon\gt 0$, there is $z$ with $d(z,y)<\epsilon$ and $d(x,z)<d(x,y)$.

Proof. If the closed ball property holds, then fix any $x,y$ with $r=d(x,y)$. Since the closure of $B_r(x)$ includes $y$, the second property follows. Conversely, if the second property holds, then if $r=d(x,y)$, then the property ensures that $y$ is in the closure of $B_r(x)$, and so the closure of the open ball includes the closed ball (and it is easy to see it does not include anything more than this, since if $g$ belongs to the closure of $B_r(x)$ then $d(x,g) \le r$ and so $g$ must also belong to the closed ball of radius $r$ centered at $x$).
QED
A: 
Let $(X,\|\cdot\|)$ be a normed linear space. Then $\overline{B_1(0)}=\bar{B}_1(0)$.

Proof. Observe that $\overline{B_1(0)}$ is the smallest closed set containing $B_1(0)$ and $B_1(0)\subset \bar{B}_1(0)$, so trivially $\overline{B_1(0)}\subset\bar{B}_1(0)$. Now to show 
$\bar{B}_1(0)\subset \overline{B_1(0)}$. Observe that, $\bar{B}_1(0)=B_1(0)\cup \partial B_1(0)$, i.e., for all $x\in \partial B_1(0), \, \exists x_n\in B_1(0)$ such that $\|x_n-x\|\to 0$: for any given $x\in \partial B_1(0),$ let $x_n=(1-\frac{1}{n})x, \, n\in \mathbb{N}.$ Then show $x_n\in B_1(0)$ and $\|x_n-x\|\to 0$. 
