How to choose the radius to show that a set is open

Question

I have the set $$ A=\{(x,y,z)\in\mathbb{R}^3:xyz\neq0\} $$

This set contains all of $\mathbb{R}^3$ except the points $(x,y,z)$ that will make $xyz=0$. So it contains

$$ \mathbb{R}^3-\{(0,y,z),(x,0,z),(x,y,0),(x,0,0),(0,y,0),(0,0,z),(0,0,0)\}\qquad\text{for all }x,y,z\in\mathbb{R}-\{0\} $$ So it does not contain the $xy$-plane, $yz$-plane, $xz$-plane, $x$ axis, $y$ axis, $z$ axis and origin. I think it is an open set.

So, I need to show that for any $x\in A$, there exists a radius $r>0$ such that the sphere $S(x,r)$ with center $x$ and radius $r$ is a subset of $A$; that is $S(x,r)\subseteq A$. I also define $S(x,r)$ with center $x$ and radius $r$ to be the set

$$ S(x,r)=\{y:|x-y|<r\} $$ The first step for me I think is to choose the radius $r$ and then show that if $x\in S(x,r)$, then $x\in A$ but I am not sure what exactly my $r$ needs to be.

Answer 1

If you have any point $(a, b, c) \in A$, then the ball $S((a, b, c), r)$ is the set of all points $(x, y, z)$ such that $$ \sqrt{(x - a)^2 + (y - b)^2 + (z - c)^2} < r, $$ or equivalently such that $$ (x - a)^2 + (y - b)^2 + (z - c)^2 < r^2. $$

We don't want to allow any points where one of $x$, $y$, or $z$ is $0$. This means, for example, that the point $(0, b, c)$ should not be in the ball, and so it must not be the case that $$ a^2 + 0^2 + 0^2 < r^2. $$

We thus need to choose $r$ such that $r \leq |a|$. Similarly, we should also choose $r$ such that $r \leq |b|$ and $r \leq |c|$. An equivalent condition is that we require that $r \leq \min(|a|, |b|, |c|)$, and in fact we can choose $r$ to be equal to $\min(|a|, |b|, |c|)$.

(Alternatively, draw yourself a picture. We want $r$ to be smaller than the distance from the centre of the ball to the $yz$-plane, and this distance is equal to $|a|$. Similarly, it must be smaller than the distances from the centre of the ball to the $xz$-plane and the $xy$-plane, which are equal to $|b|$ and $|c|$ respectively.

We now just need to confirm that this choice actually works. If $(x, y, z)$ is a point where $xyz = 0$, then one of $x$, $y$, and $z$ must be $0$. If $x = 0$, then $$ (x - a)^2 + (y - b)^2 + (z - c)^2 = a^2 + (y - b)^2 + (z - c)^2 \geq a^2 \geq r^2, $$ and so $(x, y, z)$ is not in the ball. We can handle the cases $y = 0$ and $z = 0$ in the same way.

Answer 2

Since $xyz=0$ if and only if $x=0$ or $y=0$ or $z=0$, the complement of your set is exactly

\begin{align} \mathbb{R}^3\setminus A&=\{(x,y,z)\in\mathbb{R}^3:xyz=0\}\\ &=\{(x,y,z)\in\mathbb{R}^3:x=0\}\cup\{(x,y,z)\in\mathbb{R}^3:y=0\}\cup\{(x,y,z)\in\mathbb{R}^3:z=0\}, \end{align} which is exactly the union of the $yz$, $xz$, and $xy$ planes. Each of those planes is closed, so $\mathbb{R}^3\setminus A$ is closed and hence $A$ is open.

Answer 3

If $(x,y,z) \in A$ then $x \ne 0; y \ne 0; z \ne 0$ so $|x| > 0; |y| > 0; |z| > 0$.

Let $r = \min (|x|, |y|, |z|)$.

Then if $||(x,y,z),(u,v, w)|| < r$ then $|x| \ge r > \sqrt{(x-u)^2 + (y-v)^2 + (z-w)^2} \ge \sqrt {(x - u)^2} = |x-u| $. So $|x| + |u| > |x-u| + |u| \ge |x-u + u| = |x|$. So $|u| > |x| -|x| = 0$. So $u \ne 0$.

Do the same to determine $v\ne 0; w\ne 0$. So $(u,v,w) \in A$.