How do you show the set of all vectors with a restricted norm in $\mathbb{R}^n$ is compact? 
Consider the set above. How do you show the set is bounded and closed?
I know the boundness is a given, but how does one show the set is even closed? 
The set of all vectors in space with the property that its norm is restricted to 2 and 4 can be extended to infinity because vectors are equivalent only in direction and magnitude. So wouldn't technically mean there would be a lot of vectors with this property?
I know for in the case of $\mathbb{R}$, this just dumb it down to the number line, but I am extremely lost on how to do it in $\mathbb{R^n}$
 A: Let $E$ be the set of all $x\in\mathbb R^n$ with $2\leq |x|\leq 4$. Clearly this is bounded.
Let $|x|<2$, say $|x|=2-t$. Then the open ball with centre $x$ and radius $t$ lies outside $E$. Let $|x|>4$, say $|x|=4+s$. Then the open ball with centre $x$ and radius $s$ lies outside $E$. This shows that the complement of $E$ is open so that $E$ is closed.
A: Let be $K = \{y \in \mathbb{R}^n: 2\leq \|y\| \leq 4\}$. Given $x \in \overline{K}$, there exists $(x_k)$ in $K$, such that $x_k \rightarrow x$.  Thus, for each $k$,
$$2\leq \|x_k\| \leq 4.$$
So, by continuity,
$$2\leq \|x\| \leq 4.$$
i.e., $x \in K$.
A: Let's call your set $E$. Since the norm function $N:\Bbb R^n \rightarrow \Bbb R$ is continuous, and $[2,4]$ is closed in $\Bbb R$, $N^{-1} ([2,4])=\{x \in \Bbb R^n | \  2 \leq \|x \| \leq 4\}=E$ is closed. Also, clearly $E \subset B(0,5)$, the ball of radius $5$ centered around $0$. So $E$ is bounded. By the Heine-Borel theorem, $E$ is compact.
A: One way to proceed is to show that its complement is open. Note that the complement of the set in question is the disjoint union of the open ball of all $\textbf{x}$ with $\lVert\textbf{x}\rVert<2$ and the complement of the closed ball of all $\textbf{x}$ with $\lVert\textbf{x}\rVert\geq 4$. (The latter is simply the closure of the open ball of all $\textbf{x}$ with $\lVert\textbf{x}\rVert<4$.)
A: Suppose that $x_n\to x$ and $2\leq\|x_n\|\leq4$ for all $n$. Then
$$
\|x\|\leq\|x-x_n\|+\|x_n\|\leq\|x-x_n\|+4.
$$
As $\|x-x_n\|$ can be made arbitrarily small, we get $\|x\|\leq4$.
Similarly,
$$
2\leq\|x_n\|\leq\|x_n-x\|+\|x\|,
$$
so $2\leq\|x\|$. This shows that the set contains its boundary, i.e. it is closed.
A: The norm is a continuous function from ${\mathbb R}^n$ to $\mathbb R$, and the interval $[2,4]$ is closed in $\mathbb R$.  If $C$ is a closed subset of $Y$ and $f$ is a continuous function from $X$ to $Y$, then 
$f^{-1}(C)$ is closed in $X$.
