# 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?

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).

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