Determine if the set $\{\textbf{x} \in \mathbb{R^2} : 2\leq \|\textbf{x}\| \leq 4 \}$ is compact

Question

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.

Call your set $A$. Assume $x_n\in A$ and $x_n\to x$. Then $2\leq \left|x\right|\leq4$. Now take limits. Since $\left|y\right|=\sqrt{y^2(1)+y^2(2)}$ is continuous you get that $2\leq\left|x\right|\leq 4$. Therefore, $x\in A$. — orole, Feb 10 '18 at 02:53

copper.hat · Answer 1 · 2018-02-10T04:49:30.827

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.

score 2 · Answer 2 · answered Feb 10 '18 at 02:54

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 $\epsilon$.

score 2 · Accepted Answer · answered Feb 10 '18 at 02:55

2

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 $2\leq\|x\|\leq 4$.

answered Feb 10 '18 at 02:55

user284331

56,315

Why am I free to assume the sequence converges to $\textbf{x}$? – D.C. the III Feb 10 '18 at 02:58
For a set in Euclidean to be closed, assume that $x_{n}$ is convergent and are elements of the set, and try to show that the limit is inside the set. – user284331 Feb 10 '18 at 02:59
@dc3rd, because the sequential characterization of closedness states that $F$ is closed iff for any sequence ${x_n}\subseteq F$ converging to $x$, we must have $x\in F$, i.e., any convergent sequence in $F$ must converge in $F$. – Prasun Biswas Feb 10 '18 at 03:00
You may look at this link. – user284331 Feb 10 '18 at 03:00
@user284331, what do you mean by "limit comparison rule" here? The final step seems like an application of the squeeze theorem, no? – Prasun Biswas Feb 10 '18 at 03:03
I was trying to say that if $u_{n}\leq v_{n}$ and $u_{n}\rightarrow u$, $v_{n}\rightarrow v$, then $u\leq v$, not sure what should be the appropriate labelling. – user284331 Feb 10 '18 at 03:04

score 1 · Answer 4 · answered Feb 10 '18 at 03:16

1

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.

answered Feb 10 '18 at 03:16

MPW

44,860
2
36
83

Determine if the set $\{\textbf{x} \in \mathbb{R^2} : 2\leq \|\textbf{x}\| \leq 4 \}$ is compact

4 Answers4