Show that the closed ball is closed in $\mathbb{R}^p$ 
Let $r>0, p \in \mathbb{N}$ be given. Show in detail that the closed ball $\{ x \in \mathbb{R}^p : ||x|| \leq r \}$ is closed in $\mathbb{R}^p$.

Let $A = \{ x \in \mathbb{R}^p : ||x|| \leq r \}$ for every $r>0$.
I know that we must show that $\mathbb{R}^p\setminus A$ is open in $\mathbb{R}^p$.
$\mathbb{R}^p\setminus A = \{ x \in \mathbb{R}^p: ||x|| > r \}$. 
Now let $R = ||x|| - r >0$
Am I on the right track?? I somehow think that I am not, since I cannot figure out how to show that the one is a subset of the other. Can anybody please point me in the right direction?
 A: Hint: Write the complement $A' = \mathbb R^n - A$. An arbitrary point $p \in A'$ if $||p|| = R > r$. Now see if you can find a $\delta$ such that the ball $B(p,\delta) \subset A'$. Then you will have shown $A'$ open and thus $A$ closed.
A: I do not know what $B(0,r)$ is, but you are correct in thinking that we need to show that the complement is open. To that end, let $x$ be in the complement of $A$. Then $|x|-r=\delta>0$. Let $N_x(\delta/2)$ be the set
$$
N_x(\delta/2):=\{y:|x-y|<\delta/2\}.
$$
For any $y$ in $N_x(\delta/2)$, what can you say about $|y|$? Hint, use the triangle inequality, 
$$
|x|=|x-y+y|\leq |x-y|+|y|.
$$
Once you come to the conclusion about $|y|$, you should be able to say that $N_x(\delta/2)\cap A=\emptyset$. Thus given an arbitrary $x$ in the complement of $A$, we found a neighborhood around $x$ that is contained in the complement of $A$. What can we conclude?
A: The easiest way to show that this is closed is to use sequences. Fix $p\in\mathbb N$ and $r>0$.
Let $A=\{x\in\mathbb R^p:\lVert x\rVert \leq r\}$ and pick a sequence $(a_n)$ from $A$ such that $(a_n)$ converges in $\mathbb R^p$ to the point $b\in\mathbb R^p$. To show that $A$ is closed, we need to show that $b\in A$. For each $n$, we have $\lVert a_n\rVert\leq r$ so that (by continuity of the norm) $\lVert b\rVert=\lim_{n\to\infty}\lVert a_n\rVert\leq\sup_{n\in\mathbb N}\lVert a_n\rVert\leq r$. It follows that $b\in A$, completing the proof.
To complete your proof, you need to fix a $y\in\mathbb R^p\setminus A$, and find $R>0$ such that $B(y,R)=\{z\in\mathbb R^p:\lVert z\rVert<R\}\subseteq\mathbb R^p\setminus A$. Take $R=(\lVert y\rVert - r)/2>0$. Set $B=B(y,R)$. Now show $\lVert z\rVert>r$ for each $z\in B$. (Note that $\lVert y\rVert \leq \lVert y-z\rVert + \lVert z\rVert$ and $\lVert y-z\rVert<R=(\lVert y\rVert-r)/2$.)
