Closure of Ball = Closed Ball in Normed Space

hbghlyj

Since $B(a,r)\subset\overline B(a,r)$ and a closed ball $\overline B(a,r)$ is closed(See Lemma 4.1.4.), we have$$\overline{B(a,r)}\subset\overline B(a,r)$$

---

In metric spaces, the following example will show that $\overline{B(a,r)}$ and $\overline B(a,r)$ need not equal.(MathOnline)
Consider the metric space $(M,d)$ where $M$ contains more than 1 element and $d$ is the discrete metric.
Notice that the only point that is a distance of less than 1 from $a$ is $a$ itself, so $B(a,1)=\{a\}$.
$\{a\}$ is a singleton, so it is closed(See Lemma 4.1.4.), so $\overline{B(a,1)}=\{a\}$.
Now consider the closed ball centered at $a$ with radius 1:
Every point in $M$ is of a distance of 0 or a distance of 1 from $a$, so $\overline B(a,r)=M$.
Since $M$ contains more than 1 element, we clearly see that $\overline{B(a,1)}≠\overline B(a,1)$.

---

In normed spaces, the closure of an open ball must equal the corresponding closed ball.
When is the closure of an open ball equal to the closed ball?
Closure of a Ball is Contained in Closed Ball (usual metric)
Closure of Ball = Closed Ball in Normed Space
Furthermore, in normed spaces, a closed ball is compact (Proof)

hbghlyj

A2-metricspaces.pdf
We conclude with a cautionary example which has often confused people when they first meet it.

EXAMPLE 5.1.6. In general, it need not be the case that $\bar{B}(a, \varepsilon)$ is the closure of $B(a, \varepsilon)$. Since we have seen that $\bar{B}(a, \varepsilon)$ is closed, it is always true that $\overline{B(a, \varepsilon)} \subseteq$ $\bar{B}(a, \varepsilon)$, but the containment can be proper. Indeed, take any set $X$ with at least two elements equipped with the discrete metric. Then if $x \in X$ we have $B(x, 1)=$ $\overline{B(x, 1)}=\{x\}$, but $\bar{B}(x, 1)$ is the whole space $X$.