Basically I want to prove that
$$\overline {B(\alpha; \epsilon)}=\overline B(\alpha;\epsilon)=\{x\in X: \|x-\alpha\|\leq \epsilon\}$$
First of all, we can try prove that $\overline {B(\alpha; \epsilon)} \subseteq \overline B(\alpha;\epsilon)$
let $x\in \overline {B(\alpha; \epsilon)}$, how $\overline {B(\alpha; \epsilon)}=Int(B(\alpha; \epsilon))\cup (B(\alpha; \epsilon))'$, and as an open ball in a normed space is an open in the space then $Int(B(\alpha; \epsilon))=B(\alpha; \epsilon)$, and how $B(\alpha; \epsilon)\subseteq \overline B(\alpha;\epsilon)$ then $x \in \overline B(\alpha;\epsilon)$
If $x\in (B(\alpha; \epsilon))'$, as every point of accumulation is of adherence, the result is also obtained. So $\overline {B(\alpha; \epsilon)} \subseteq \overline B(\alpha;\epsilon)$.
Now, i will try show that $\overline B(\alpha;\epsilon)\subseteq \overline {B(\alpha; \epsilon)}$.
Let $x\in \overline B(\alpha;\epsilon)$ then $| |x-\alpha| |\leq \epsilon$, we can verify that $x$ is the accumulation point in the open ball and it would already be.
For this purpose, we define the sequence $\{x_{n}\}=(1-\frac{1}{n})x+\frac{1}{n}\alpha$, note that
$$0\leq \|x_{n}-x\|=\left\|\left(1-\frac{1}{n}\right)x+\left(\frac{1}{n}\right)\alpha-x \right\|\leq \frac{\epsilon}{n} \rightarrow 0$$
This happens when $n \rightarrow \infty$, so the sequence converges but it remains to be seen if the sequence belongs to the open ball, so let's try that
$$\|x_{n}-\alpha\| = \left\|\left(1-\frac{1}{n}\right)(x-\alpha) \right\|\leq \left(1-\frac{1}{n}\right)\|x-\alpha\| \leq \left(1-\frac{1}{n}\right) \epsilon < \epsilon$$
So it seems that the sequence belongs to space, with it $x$ would be an accumulation point in the open ball and the result would be done
can this work? is there something to improve?
Another question I have is, if I consider the sequence
$\{x_{n}\}=\alpha+\left(1-\frac{1}{n}\right)(x-\alpha)$ It seems that this sequence converges to $x$, but I don't know how to guarantee that it belongs to $B(\alpha;\epsilon)$
Thanks and sorry if I omitted too many details.
