Consider the set $A = \{ (x, y, z) :\, z\leq 6 \}\subseteq \mathbb R^3$. Show that $A$ is closed.

Question

Consider the set $A = \{ (x, y, z) :\, z\leq 6 \}\subseteq \mathbb R^3$. Show that $A$ is closed.

Answer 1

If $(x,y,z)$ is a point with $z>6$ then the open ball of radius $\epsilon$ around it is contained in $A$ as long s $\epsilon <z-6$. This is because $|z'-z| <z-6$ implies $z' >6$. [$z' \geq z-|z-z'| >z-(z-6)=6$].

Answer 2

Well, the complement of $A$ is given by $\Bbb{R}^3 \setminus A = \{ (x,y,z)\in \Bbb{R}^3:\, z>6 \}$. Now, we claim that $\Bbb{R}^3 \setminus A$ is an open set, that is, for every point $x=(x_1,x_2,x_3) \in \Bbb{R}^3 \setminus A$ there exists an $r>0$ such that $B_r(x) \subset \Bbb{R}^3 \setminus A$.

Then, consider $r:=\frac12 (x_3-6)>0$, and show that $B_r(x) \subset \Bbb{R}^3 \setminus A$. For this, set an arbitrary point $(y_1,y_2,y_3) \in B_r(x)$ and then show that $y_3>6$.