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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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    $\begingroup$ In the Euclidean metric space $R^n$ it is necessarily true. $\endgroup$ – Shahab Feb 11 '12 at 2:26
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ The property may fail for subspaces of Euclidean space. See here: link link $\endgroup$ – Dejan Govc Feb 11 '12 at 11:53
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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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  • $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ 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
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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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  • $\begingroup$ 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
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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 $B[x_0;R]\subseteq\overline{B(x_0;R)}^{\|\cdot\|}$.

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