How do I link dimension of a normed vector space with closedness? Let $X$ be a Normed Vector Space, for any $x\in X$ and $r>0$. Let $W:=\{y\in X : \|y-x\|\leq r\}$ and $S:=\{y\in X : \|y-x\|<r\}$
Prove:
$W$ is closed if $\dim(X)<\infty$
I can't think of a reliable way to link dimension with closedness, perhaps im missing a theorem?
 A: The closed ball is always closed in $X$ independent of the dimension. Let $x_0\in W$, $r>0$ and $W$ as above. Let $(x_n)_{n \in \mathbb{N}} \subseteq W$ which converges in $X$ to $l\in W$. To verify the closedness of $W$ we need to check that $l\in W$. As $x_n \rightarrow l$ we can choose for every $n\in \mathbb{N}_{\geq1}$ a $m_n\in \mathbb{N}$ such that
$$ \Vert l - x_{m_n} \Vert \leq \frac{1}{n}.$$
Using the triangle inequality we get
$$ \Vert l - x \Vert \leq \Vert l -x_{m_n} \Vert + \Vert x_{m_n} - x \Vert \leq \frac{1}{n} + r. $$
As it holds true for every $n \in \mathbb{N}_{\geq 1}$, we conclude
$$ \Vert l - x \Vert \leq r $$
and therefore $l\in W$. As we have taken an arbitrary sequence, we get that $W$ is closed.
Note that the same argument yields for an metric space $(X,d)$ that $\{y\in X: d(x,y)\leq r \}$ is closed in $X$.
Edit: Furthermore, one has in normed spaces $\overline{S}=W$.
From basic topology we know that for any $A, B\subseteq X$ with $A\subseteq$ holds $\overline{A} \subseteq \overline{B}$. Thus we get
$$ S\subseteq W \Rightarrow \overline{S} \subseteq \overline{W}=W. $$
Where we used that closed sets coincide with their closure.
We are left to prove the converse inclusion. Clearly $x\in S$. Let $y\in W\setminus \{ x \}$. Define 
$$ y_n := \frac{1}{n} x + \left( 1 - \frac{1}{n} \right) y.$$
We compute
$$ \Vert y_n -x \Vert \leq \frac{1}{n} \underbrace{\Vert x - y \Vert}_{>0, \text{ as } y\neq x} + \Vert y-x \Vert  < \Vert y -x \Vert \leq r.$$
Hence, $\Vert y_n - x \Vert < r$ and thus $y_n\in S$. We compute
$$ \Vert y - y_n \Vert = \frac{1}{n} \Vert x -y \Vert \rightarrow 0, \quad n \rightarrow \infty.$$
This means $y_n \rightarrow y$. But as $(y_n)_{n\in \mathbb{N}}\subseteq S$ we get $y\in \overline{S}$. As $y\in W$ arbitrary, this yields $W \subseteq \overline{S}$.
