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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 $\mathbb{R^n}$

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"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." The interpretation of a vector as a thing with direction and magnitude, and which can be otherwise moved around without changing the vector, is not helpful for this problem. There is a precise definition of what $\mathbb R^n$ means, and in this context a vector is simply an element of (or "point" in) $\mathbb R^n$. Roughly, you should be picturing the region between two spheres, including the spheres themselves. –  Jonas Meyer Nov 1 '12 at 5:20
    
@JonasMeyer, that's what I thought too. It's some kind of spherical annulus here. I am not sure how to explain it though –  Hawk Nov 1 '12 at 5:22
    
Actually why is the "direction and magnitude" thing wrong in the first place? Other than the fact we view vectors as "points" here. Why is the reasoning wrong? –  Hawk Nov 1 '12 at 5:29
    
What reasoning? I was addressing what seemed to be misguided intuition about the set of vectors in question, that didn't describe the actual set. E.g., in $\mathbb R^1$ this is the set $[-4,-2]\cup[2,4]$, not a set of arrows pointing left or right with magnitude between 2 and 4. While in $\mathbb R^n$ it is possible to interpret a point as a "position vector" pointing from the origin to the actual point, that is not helpful here, and the statement that these vectors "can be extended to infinity" indicates something other than the set in question. –  Jonas Meyer Nov 1 '12 at 5:33
    
Could you illustrate the same example as you did for $\mathbb{R^1}$, but for $\mathbb{R^2}$? –  Hawk Nov 1 '12 at 5:48

6 Answers 6

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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., $x \in K$.

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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.

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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.

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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 $\lVert\textbf{x}\rVert<4$.)

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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.

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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 $X$.

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