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?

  • 2
    $\begingroup$ In the Euclidean metric space $R^n$ it is necessarily true. $\endgroup$ – Shahab Feb 11 '12 at 2:26
  • 5
    $\begingroup$ 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. $\endgroup$ – Alex Lapanowski Feb 11 '12 at 2:28
  • 3
    $\begingroup$ 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 $y=\lim\limits_{t\searrow 0}tx+(1-t)y$. $\endgroup$ – Jonas Meyer Feb 11 '12 at 3:37
  • 1
    $\begingroup$ The property may fail for subspaces of Euclidean space. See here: link link $\endgroup$ – Dejan Govc Feb 11 '12 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 $d(x,z)<d(x,y)$.

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

  • $\begingroup$ Thanks for the response! Your proof makes sense. $\endgroup$ – Alex Lapanowski Feb 11 '12 at 3:10
  • $\begingroup$ How to understand what is not working initially in order to build your characterization? In other words, what is the intuition behind this? Thanks. $\endgroup$ – user169373 Sep 23 '14 at 20:41
  • 3
    $\begingroup$ @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. $\endgroup$ – JDH Sep 23 '14 at 21:10
  • $\begingroup$ @JDH Sorry to ask a question for such an "old" post, but I don't understand why "since the closure of $B_r (x)$ includes $y$, the second property follows". Why $y$ being in the closure of $B_r (x)$ should mean that there is a $z$ such that $d(z,y) < \epsilon$ and $d(x,z) < d(x,y)$? We only know that $y$ is in the closure of $B_r (x)$, but how does that ensure that there are others elements for which the second property holds? And, one more thing: if $z=y$, then $d(z,y) = d(y,y) = 0 < \epsilon$ (so it works), but $d(x,z)=d(x,y)<d(x,y)$ (it doesn't work). What am I missing? Thanks in advance. $\endgroup$ – justdoit Apr 26 '17 at 8:50
  • 1
    $\begingroup$ 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. $\endgroup$ – JDH Apr 26 '17 at 11:03

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

  • $\begingroup$ How would you suggest approaching this, showing that the interior of the closed ball is equal to the open ball? $\endgroup$ – kathystehl Jan 31 '18 at 0:15

More general, we can prove that for any normed space $(V,\|\cdot\|)$, if $x_0\in V$ and $R>0$, then $$\overline{B(x_0;R)}^{\|\cdot\|}=B[x_0;R],$$ were $\overline{B(x_0;R)}^{\|\cdot\|}$ 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 $B[x_0;R]\subseteq\overline{B(x_0;R)}^{\|\cdot\|}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.