What is the closure of an open ball $B_X(\mathbf{a},r)$ in $X=\mathbb{R}^n$? Suppose we have the open ball $B_{X}(\mathbf{a},r)$ and the closed ball $\bar{B}_{X}(\mathbf{a},r)$ of radius $r$ about $\mathbf{a}\in\mathbb{R}^n=X$ with the Euclidean metric $d_2$.
What is the closure of an open ball $B_{X}(\mathbf{a},r)$ in $(\mathbb{R}^n,d_2)$?
First, I have defined the sets like this:
$$B_X(\mathbf{a},r)=\{\mathbf{y}\in\mathbb{R}^n:d_2(\mathbf{a},\mathbf{y})<r\}$$
$$\bar{B}_X(\mathbf{a},r)=\{\mathbf{y}\in\mathbb{R}^n:d_2(\mathbf{a},\mathbf{y})\le r\}$$
Then, let $U=B_X(\mathbf{a},r)$ and claim the closure is $$\bar{U}=\{\mathbf{x}\in\mathbb{R}^n:\exists (\mathbf{x}_n)\in U\text{s.t}~\mathbf{x}_n\to \mathbf{x}\}=\bar{B}_X(\mathbf{a},r).$$
Since closed balls are closed then we have $$\bar{U}\subset\bar{B}_X(\mathbf{a},r).$$
But, how can I prove the other direction $\bar{B}_X(\mathbf{a},r)\subset\bar{U}$?
Since it's a containment argument I know that I have to take any element from $\bar{B}_X(\mathbf{a},r)$ and show it is in $\bar{U}=\{\mathbf{x}\in\mathbb{R}^n:\exists (\mathbf{x}_n)\in U\text{s.t}~\mathbf{x}_n\to \mathbf{x}\}$.
But I do not know how to get an expression for $\mathbf{x}_n$.
I would appreciate any help.
 A: To show $\overline U \subseteq \overline B(\mathbf a,r)$ it suffices to show
$$\mathbf x\notin \overline B(\mathbf a,r) \implies \mathbf x\notin \overline U.$$
If $\mathbf x\notin \overline B(\mathbf a,r)$ then we have $d(\mathbf a,\mathbf x)>r$. 
If we choose $0<\delta<d(\mathbf a,\mathbf x)-r$ then we can show (using triangle inequality) that 
$$B(\mathbf x,\delta) \cap B(\mathbf a,r) = \emptyset.$$
This means that $\mathbf x$ has a neighborhood which does not intersect $U$ and, consequently, $\mathbf x$ does not belong to the closure of $U$.
A different argument to see this inclusion: Since the closed ball $\overline B(\mathbf a,r)$ is closed set and $U\subseteq\overline B(\mathbf a,r)$, we also have $\overline U\subseteq\overline B(\mathbf a,r)$.

Now let us have a look at $\overline B(\mathbf a,r)\subseteq \overline U$, i.e. that every point of the closed ball belongs to the closure.
If $d(\mathbf a,\mathbf x)<r$, then $x\in U$ and thus $x\in\overline U$. It only remains to look at the points such that $d(\mathbf x,\mathbf a)=r$.
Notice that for any $\alpha$ we can find a point $\mathbf p_\alpha= \alpha \mathbf a + (1-\alpha) \mathbf x$ such that $d(\mathbf a,\mathbf p_\alpha)=(1-\alpha)r$ and $d(\mathbf x,\mathbf p_\alpha)=\alpha r$. (This is the first time we are using that we work with the Euclidean metric. It might also be helpful if you try to draw a picture. You should see that for $\alpha\in(0,1)$ we get exactly the points on the straight lines between $\mathbf a$ and $\mathbf x$.)
Indeed, if the $i$-th coordinate of $\mathbf p_\alpha$ is $p_i=\alpha a_i+(1-\alpha) x_i$
$$d(\mathbf x,\mathbf p_\alpha) = \sqrt{\sum_{i=1}^n (p_i-x_i)^2} = \sqrt{\sum_{i=1}^n \alpha^2(a_i-x_i)^2} = \alpha \sqrt{\sum_{i=1}^n (a_i-x_i)^2} = \alpha d(\mathbf x,\mathbf a).$$
The other equality is shown similarly.
So if we take $\mathbf x_n=\mathbf p_{\frac1n}$, then this is a sequence of points from $U$ which converges to $\mathbf x$.
Maybe it is useful to mention that this proof make become a bit clearer if you rewrite it using the norm $\|\mathbf v\|_2=\sqrt{v_1^2+\dots+v_n^2}$ rather than the metric $d_2(\mathbf x,\mathbf a)=\|\mathbf x-\mathbf a\|_2$. (But I wrote this down using metric, since I do not know whether you are familiar with the notion of norm.) 
