Determine if the set $\{\textbf{x} \in \mathbb{R^2} : 2\leq \|\textbf{x}\| \leq 4 \}$ is compact Determine if the set $\{\textbf{x} \in \mathbb{R^2} : 2\leq \|\textbf{x}\| \leq 4 \}$ is compact
Defn: A set is closed if it contains all of its limit points
Defn: A set is bounded if $\exists$ R such that the set $A$ is contained in the $B_{R}(0)$ 
Attempt:
I say this set is compact. But I don't know how to show it is closed.
Proof for bounded: $$\forall \textbf{x} \in \mathbb{R^n} \exists M$$ s.t $$\|\textbf{x}\| \leq M$$ If I am going strictly by the definition of bounded in $\mathbb{R^n}$ then I could let $M = 4$ and I should be fine. 
Proof for Closed Set:
This is where I am stuck. 
Note: I am aware of being able to use continuous functions as a way to show a set is closed, but I am working through my textbook and up to this point we have not covered continuous functions. So I was looking for a method that did not perhaps include that notion.
 A: The function $f(x) = \|x\|$ is continuous, and so $f^{-1} ( [2,4])$ is closed.
Alternatively, the set $[2,4] \times [0, 2 \pi]$ is compact, and the
function $g((r,\theta)) = (r \cos \theta, r \sin \theta)$ is continuous
hence $g([2,4] \times [0, 2 \pi]) $ is compact.
Here is a sequential way: Let $C$ be the set in question.
Note that the triangle rule gives $| \|x\| - \|y\| | \le \|x-y\|$, so if we
have sequence $x_n \to x$ then we have $\|x_n\| \to \|x\|$. The set $C$ is closed
if whenever $x_n \to x$ with $x_n \in C$, then $x \in C$.
So, suppose $x_n \in C$ and $x_n \to x$. We must show that $x \in C$, or equivalently, $2 \le \|x\| \le 4$. Since $x_n \in C$ we have
$2 \le \|x_n\| \le 4$ and since $\|x_n\| \to \|x\|$, we have $2 \le \|x\| \le 4$ and so $x \in C$. Hence $C$ is closed.
A: If one of its limit points is outside the set, then there will be contradiction because there are some $\epsilon >0$ such that no point of the set lies in the ball of the limit point with radius $\epsilon$.
A: Assume that $\{x_{n}\}$ is such that $x_{n}\rightarrow x$ and $2\leq \|x_{n}\|\leq 4$, then $\|x_{n}\|\rightarrow\|x\|$, then by Limit Comparison Rule we have $2\leq\|x\|\leq 4$.
A: This set is of the form $K\setminus U$, where $K$ is the closed ball of radius $4$ centered at $0$ and $U$ is the open ball of radius $2$ centered at $0$. $K$ is closed and bounded, and $U$ is open. Done.
