# 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?

• In the Euclidean metric space $R^n$ it is necessarily true.
– user10575
Commented Feb 11, 2012 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. Commented Feb 11, 2012 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 $y=\lim\limits_{t\searrow 0}tx+(1-t)y$. Commented Feb 11, 2012 at 3:37
• The property may fail for subspaces of Euclidean space. See here: link link Commented Feb 11, 2012 at 11:53

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

• Thanks for the response! Your proof makes sense. Commented Feb 11, 2012 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
Commented Sep 23, 2014 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
Commented Sep 23, 2014 at 21:10
• The way that $y$ gets into the closure of $B_r(x)$ is that there must be points in $B_r(x)$ that are as close as you like to $y$. Those points are the $z$'s in the property.
– JDH
Commented Apr 26, 2017 at 11:03
• @James Well since y is in closure of B(x,r), for any $\epsilon>0$, $B(y,\epsilon) \cap B(x,r) \neq \phi$. Choose z from this intersection. Then, $d(x,z)<r=d(x,y)$ Commented Jan 14, 2021 at 11:40

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

• How would you suggest approaching this, showing that the interior of the closed ball is equal to the open ball? Commented Jan 31, 2018 at 0:15
• ... where a "normed linear space" means over some field with an archimedean valuation. The result is not true e.g. over $p$-adics where that $1-1/n$-construction fails. -- Also, note that $\delta B_1(0)$ here should really be defined as $\{x \in X: \lvert \lvert x \rvert \rvert =1 \}$, as only a posteriori we show this is the boundary also in the topological sense. Commented Apr 11, 2020 at 7:14

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 $$\begin{eqnarray} \|x_n-x_0\| & = & \left\|\left(1-\frac{1}{n}\right)x+\frac{1}{n}x_0-x_0\right\| \\ & = & \left\|\left(1-\frac{1}{n}\right)x-x_0\left(1-\frac{1}{n}\right)\right\| \\ & = & \left\|\left(1-\frac{1}{n}\right)(x-x_0)\right\| \\ & = & \left(1-\frac{1}{n}\right)\underbrace{\|x-x_0\|}_{\leqslant R} \\ & \leqslant & \underbrace{\left(1-\frac{1}{n}\right)}_{<1}R which gives us $$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 $$B[x_0;R]\subseteq\overline{B(x_0;R)}^{\|\cdot\|}$$.

• Cf. my comment to Lyapunav's answer. Commented Apr 11, 2020 at 7:15
• @TorstenSchoeneberg if you need this, I have no problem to give you credits... Commented Apr 11, 2020 at 12:34
• I think what Torsten meant was that this isn't a proof for a general normed space. 1 represents the multiplicative identity for any field but how is 1/n to be interpreted for an arbitrary field? Commented Apr 26, 2022 at 3:11