Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am having trouble with these two questions. In particular, using a ball and choosing an $r$ to show that a set is open.

(a) $$X = \left \{ \mathbf{x} \in \mathbb{R}^d | \: ||\mathbf{x}|| \leq 1 \right \} .$$

So, $X$ is closed if its compliment $X^c$ is open. So if I can show that $X^c$ is open, then it follows that $X$ is closed. I'll start with the definition I am using for a ball.

The ball about $\mathbf{a}$ in $\mathbb{R}^n$ of radius $r$ is the set $$B_{r}(\mathbf{a})= \left \{ \mathbf{x} \in \mathbb{R}^n : || \mathbf{x} - \mathbf{a} || < r\right \}$$ A subset $U$ of $\mathbb{R}^n$ is open if for every $\mathbf{a} \in U$, there is some $r=r(a) > 0$ such that the ball is contained in $U$.

So, I have a set $X^{c} = \left \{ \mathbf{v} \in \mathbb{R}^d | \: ||\mathbf{v}|| > 1 \right \} $. It consists of points whose lengths are longer than 1. Let $\mathbf{a} \in X^{c}$, then $|| \mathbf{a} || > 1$. Now, using $B_{r}(\mathbf{a})$ as it is written above, how do I find an explicit formula for $r>0$? Intuitively it seems to make sense if $\mathbf{x}$ is in $X$, then $0<|| \mathbf{x} - \mathbf{a} ||$ as $||\mathbf{x}|| \neq ||\mathbf{a}||$. I am not sure how to proceed.

(b) $$X = \mathbb{R}^2 \setminus \left \{ \mathbf{x} \in \mathbb{R}^2 | \mathbf{x}=(x,0) \: \right \}$$

I know that this set is open, but it again comes down to choosing some $r$ and using a ball. The set is $\mathbb{R}^2$ less a line across the x-axis - that is, any $x$ and any $y$ with $y \neq 0$. The line has no height, so it should be easy to show that for any $\mathbf{a} \in X$ there exists $r>0$ such that $B_r(a) \subseteq X$.

I've done lots of scratch work and diagrams etc. but I just can't seem to put this concept together. Any help and clarification would be appreciated.

share|improve this question

3 Answers 3

up vote 1 down vote accepted

HINT: Suppose that $\mathbf{x}\in\Bbb R^d$ with $\|\mathbf{x}\|=r>1$. Let $\epsilon=r-1$. Does $B_\epsilon(\mathbf{x})$ contain any points of the closed unit ball $X$? (If you’re not sure about that $\epsilon$, cut it in half.)

Added: Sorry: I missed the second question originally. You can use the same idea. Suppose that $\mathbf{p}=\langle x,y\rangle\in X$, so that $y\ne 0$. Let $r=|y|$; does $B_r(\mathbf{p})$ contain any point of the $x$-axis?

share|improve this answer

Regarding you second question: Let $a\in X$, and you want to find $r>0$ such that $B_r(a)\subseteq X$. What could "go wrong"? Well, this ball might contain points of the form $(x,0)$, and these are not in $X$. So all we have to do is to eliminate this option. Try to figure out a general way of doing it, using an example. Suppose $a=(1,8)$. If you take $r=9$ it won't work, right? But it will work for all $r<8$.

share|improve this answer

For (a) draw a plane figure. Choose a point $a\notin X$. How large a circle can you draw around $a$ such that its interior does not intersect $X\>$? Use some form of the triangle inequality to show that what you see is indeed true.

With (b) it's the same thing. Choose a point $a:=(a_1,a_2)\notin X$. How large a circle can you draw around $a$ such that its interior does not intersect $X\>$?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.