The Set $\{x\in \mathbb{R}^n : \|x\|\leq r\}$ is Closed How to show that a set $A=\{x\in \mathbb{R}^n : \|x\|\leq r\}$ is closed? 
May I do it showing that for $x\in A, a\in \mathbb{R}$ it holds $ax\in A$ and for $x,y\in A$ it holds $x+y \in A$?
 A: Here closed means that $\partial A$= boundary of A belongs to $A$.
And set theoretically $\partial A=\{x\in R^n|\forall r>0,B(x,r)\cap A\ne \phi \text{ and }B(x,r)\cap A^c\ne \phi\}$
$B(x,r)=\{y\in R^n| ||(y-x)||=d(y,x)<r\}$ is the open ball around $x$ of radius $r$.
Now to show this,
Let $y\in \partial A$.I will show that $y\in A$.
Let if possible $y\notin A\Rightarrow y\in A^c$
Then $\forall r>0,y\in B(y,r)\cap A^c$ so $B(y,r)\cap A^c\ne \phi$
As $y\notin A$ so $\exists \{y_n\}_{n\ge 1}$ with $y_i\in A$ and $y_i\to y$ as $n\to \infty$
But this implies $\forall \epsilon>0,\exists n\in N\text{ such that }||(y_i-y)||<\epsilon\text{ for all }i\ge n$
We know as $y_i\in A,||x-y_i||\le r\Rightarrow ||x-y||=||x-y_i+y_i+y||\le ||(x-y_i)||+||(y_i-y)||\le r+\epsilon$
As $\epsilon>0$ was arbitrary this implies $||(x-y)||\le r$ but this means $y\in A$ which contradicts our hypothesis. So $y\notin A$ is not possible.
So we have $y\in A$ and as $y\in \partial A$ was arbitrary this implies $\partial A \subseteq A$. Hence $A$ is closed.
A: You are confusing here between two meanings of the word closed. The set you consider is not closed in the meaning you are interpreting. In fact, neither of the properties you mention is true in this case. 
Closed in this context is topological. You need to show that the set contains all of its accumulation points. So, given a sequence of points $x_n$ with $\|x_n\|\le r$, you need to show that if $y$ is the limit of the sequence, then $\|y\|\le r$.
A: One way to show a set is (topologically) closed is to show that the complement of that set is open. To that end, pick any $y \in A^c$. By virtue of the fact that $y \notin A$, it must be that $||y|| > r$. In other words, we have $||y|| = r + s$ for some $s > 0$.
Now, the set
$$
\left \{x \in \mathbb{R}^n : \|x-y\| < \frac{s}{2} \right \}
$$
is open and contained within $A^c$. As $y$ was chosen arbitrarily, it follows that every point of $A^c$ has an open neighborhood contained within $A^c$, and therefore $A^c$ is open.
