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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?

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In the Euclidean metric space $R^n$ it is necessarily true. – Shahab Feb 11 '12 at 2:26
Right, but Euclidean space is known for, among other things, being perfect in almost every way. What about spaces like $L^{p}$ or $H^{p}$? I'm looking to see how far our intuition of Euclidean spaces and the standard metric extends. – Alex Lapanowski Feb 11 '12 at 2:28
Regarding your question about $L^p$ or $H^p$, it is true in every normed space. If $\|x-y\|=r$, then for $0<t<1$, $\|x-(tx+(1-t)y)\|=(1-t)\|x-y\|<r$, and $y=\lim\limits_{t\searrow 0}tx+(1-t)y$. – Jonas Meyer Feb 11 '12 at 3:37
The property may fail for subspaces of Euclidean space. See here: link link – Dejan Govc Feb 11 '12 at 11:53
up vote 18 down vote accepted

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. QED

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Thanks for the response! Your proof makes sense. – Alex Lapanowski Feb 11 '12 at 3:10
How to understand what is not working initially in order to build your characterization? In other words, what is the intuition behind this? Thanks. – user169373 Sep 23 '14 at 20:41
@MarcGato Think about the case where you have an isolated point $x$, so that $B_r(x)$ contains only $x$, for some $r>0$, but there is a point $y$ at distance $r$ to $x$. The closure of $B_r(x)$ is just $\{x\}$, since there are no other limit points to add and so this set is already closed. My condition ensures that every point at distance $r$ from $x$ is a limit of points of distance less than $r$ from $x$. That is why all such points get added to the closure of the open ball. – JDH Sep 23 '14 at 21:10

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 $\|x_n-x\|\to 0$.

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