Is there any metric space $(X, d)$ other than the particular class normed space where $\overline {B(x, r)}=B[x, r]$ for all $x\in X, r>0$?

Question

$(X, d) $ be a metric space.

$a\in X, r>0$

$$B(a, r) =\{x\in X : d(a, x) <r \}$$

$$B[a, r]=\{x \in X : d(a, x) \le r \}$$

$\overline{B(a,r)}=Closure[B(a,r)]$

Question : When closure of an open ball $B(a, r)$ is the closed ball $B(a, r) $?

1. $B(a, r) \subset B[a, r]$

And $\overline {B(a, r) } $ is the smallest closed set containing $B(a, r) $ .

Hence, $\overline {B(a, r) } \subseteq B[a, r]$

Now,the question is when $B[a, r]\subseteq \overline {B(a, r) }$?

2. In general in a metric space the above set inclusion may not be true.

Example: $(X, d) $ be a discrete space with $\text{ } card(X) \ge 2$.

Then, $\overline{B(a, 1)} =\overline{ \{a\}}=\{a\}\neq X =B[a, 1]$

3. $(X, ||•||) $ be a normed space then $\overline {B(0, 1)} =B[0, 1]$

Proof: It is enough to show $B[0, 1]\subseteq \overline {B(0, 1)}$

Let, $x\in B[0, 1] $ to show $x \in \overline {B(0, 1)}$

Choose a scalar sequence $ \lambda _n = \frac{n}{n+1}$

And set, $x_n = (\lambda_n) x$

Then, $||x_n||=|\lambda_n |•||x||<1$

And $(\lambda_n x )$ converges to $ x$ in $(X,||•||)$ [as $\lambda_n \to 1$]

Hence, we proved the existence of a sequence $(x_n)_{n\in {\mathbb{N}} }\subset B(0,1)$ such that $(x_n) \to x$ implies $x\in \overline {B(0, 1)}$.

Hence,$\overline {B(0, 1)}=B[0, 1]$

Question:

Is there any metric space $(X, d)$ other than the particular class normed space where $\overline {B(x, r)}=B[x, r]$ for all $x\in X, r>0$ ?

Thanks. Forgive me for any mistake.

Answer 1

No, the metric $d$ is not necessarily induced by a norm. For example, if $(X,\lVert\cdot\rVert)$ is a normed space, then the metric $d$ given by $d(x,y)=\|x-y\|^{1/2}$ is certainly not induced by a norm (the scaling is wrong), but clearly $\overline{B(x,r)}=B[x,r]$ for all $x\in X$ and $r>0$.

Another example is provided by Riemannian metrics. If $g_x$, $x\in \mathbb R^n$, is an inner product that depends continuously on $x$, then $$ d(x,y)=\inf\left\{\int_0^1g_{\gamma(t)}(\dot\gamma(t),\dot\gamma(t)^{1/2}\,dt:\gamma\in C^\infty([0,1];X),\,\gamma(0)=x,\gamma(1)=y\right\} $$ defines a metric with the desired property, but it is rarely induced by a norm (for example, if you scale $g_x$ to become sufficiently small as $\|x\|$ gets big, you can get a metric with finite diameter).