Consider the Metric space $(\mathbb{R}^n,\|\cdot\|)$. Show that $Cl(B(x,r))=C(x,r)$ with $C(x,r)$ the closed ball around $x\in X$ with radius $r$.
$\operatorname{Cl}(B(x,r))\subseteq C(x,r)$ is trivially true (the closure is the smallest closed set containing $B(x,r)$).
I, however, have difficulties proving it the other way around. I am able to prove it when I introduce the concept of a boundary point and show that $\operatorname{Cl}(B(x,r))=B(x,r) \cup \partial(B(x,r))$, but the concept of a boundary hasn't been introduced yet in the lectures. Can anyone give me a hint on how to prove this without using the concept of a boundary?
Definition: Let $x\in X$. We say $x$ is a closure point of $A\subseteq X$ when $\forall r>0$ $B(x,r)\cap A\neq\emptyset$. We denote the set of all closure points as $\operatorname{Cl}(A)$
