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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.

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up vote 2 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?

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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 $r<8$.

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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 $X\>$?

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