$(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 $B(a, r)$ in a metric space $(X,d)$ is the closed ball $B(a, r) $?
My progress:
$B(a, r) \subset B[a, r]$
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 $B[a, r]\subseteq \overline {B(a, r) }$?
- 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, $\overline{B(a, 1)} =\overline{ \{a\}}=\{a\}\neq X =B[a, 1]$
- $(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,$\overline {B(0, 1)}=B[0, 1]$
- I have posted this question before. From there i came to know about a linear metric space $(X,d)$ where the metric doesn't induced by norm but closure of every open ball $B(a, r) $ is the closed ball $B[a, r]$ for all $x\in X, r>0$.
Here is the example of the space
Question:
When the closure of the open ball $\overline{B(a,r)}=B[a,r]$ forall $a\in X, r>0$ in a metric space $(X, d)$ ?
Is there any way to classify all metric spaces where the closure of any open ball is the closed ball with same center and same radius?
Thanks. Forgive me for any mistake.
