How to show $S={{(x,y,0) : x,y \in \Bbb{R}}}$ is closed? Let $V= \Bbb{R}^3$ and $S=\{(x,y,0) : x,y \in \mathbb{R}\}$ 
Show that $S$ is closed. I understand that I need to show that $S^\complement$ is open but that is as far as I can go. We've been learning how to do this with open balls $B_r(x)$ 
 A: Let $(x,y,z) \in S^\complement$. Note that $z \neq 0$ and that $B_{|z|}(x,y,z)$ contains no points where the $z$-component is zero, so $B_{|z|}(x,y,z)$ has empty intersection with $S$.  Hence, $B_{|z|}(x,y,z) \subset S^\complement$. Since this can be done for every point $(x,y,z) \in S^\complement$, it follows that $S^\complement$ is a union of open balls, so it is open.  Thus, $S=(S^\complement)^\complement$ is closed.
A: Pick a point
$p = (x, y, z) \in \Bbb R^3 \setminus S; \tag 1$
then
$z \ne 0; \tag 2$
set
$\epsilon = \dfrac{\vert z \vert}{2} > 0; \tag 3$
let
$B_\epsilon(p) = \{q \in \Bbb R^3 \mid \vert q - p \vert < \epsilon \}; \tag 4$
that is, $B_\epsilon(p)$ is the open ball of radius $\epsilon$ about $p$; observe that
$B_\epsilon(p) \subset S^c, \tag 5$
since the radius of the ball is half the distance 'twixt its center $p$ and $S$; since such considerations apply to any $p \in S^c$, that is, every such $p$ is contained in an open ball contained in $S^c$, we conclude that $S^c$ is open and hence that $S$ is closed in $\Bbb R^3$.
A: With a bit more theory at one's disposal, one could simply do the following:
The function $\pi_3:\mathbb{R}^3 \to \mathbb{R}: (x,y,z) \to z$ is continuous. And $S=\pi_3^{-1}[\{0\}]$, the inverse image of a closed set under a continuous function, hence closed.
