96
$\begingroup$

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?

$\endgroup$
4
  • 4
    $\begingroup$ In the Euclidean metric space $R^n$ it is necessarily true. $\endgroup$
    – Shahab
    Feb 11, 2012 at 2:26
  • 6
    $\begingroup$ 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. $\endgroup$ Feb 11, 2012 at 2:28
  • 4
    $\begingroup$ 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 $y=\lim\limits_{t\searrow 0}tx+(1-t)y$. $\endgroup$ Feb 11, 2012 at 3:37
  • 3
    $\begingroup$ The property may fail for subspaces of Euclidean space. See here: link link $\endgroup$
    – Dejan Govc
    Feb 11, 2012 at 11:53

3 Answers 3

62
$\begingroup$

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

$\endgroup$
8
  • $\begingroup$ Thanks for the response! Your proof makes sense. $\endgroup$ Feb 11, 2012 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, 2014 at 20:41
  • 7
    $\begingroup$ @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, 2014 at 21:10
  • 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, 2017 at 11:03
  • 1
    $\begingroup$ @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, $d(x,z)<r=d(x,y)$ $\endgroup$
    – Believer
    Jan 14, 2021 at 11:40
21
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ How would you suggest approaching this, showing that the interior of the closed ball is equal to the open ball? $\endgroup$
    – kathystehl
    Jan 31, 2018 at 0:15
  • 2
    $\begingroup$ ... 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. $\endgroup$ Apr 11, 2020 at 7:14
7
$\begingroup$

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 $B[x_0;R]\subseteq\overline{B(x_0;R)}^{\|\cdot\|}$.

$\endgroup$
3
  • 2
    $\begingroup$ Cf. my comment to Lyapunav's answer. $\endgroup$ Apr 11, 2020 at 7:15
  • $\begingroup$ @TorstenSchoeneberg if you need this, I have no problem to give you credits... $\endgroup$
    – Mathecm
    Apr 11, 2020 at 12:34
  • $\begingroup$ 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? $\endgroup$
    – ngc1300
    Apr 26 at 3:11

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.