Is the set $S:=\{(x_1,x_2)\in \mathbb{R}^2: x_2>|x_1|\}$ open, or closed or neither? Is the set $S:=\{(x_1,x_2)\in \mathbb{R}^2: x_2>|x_1|\}$ open, or closed or neither?
Intuitively, and graphically it is obvious, that I can draw an open ball for any $x \in S$, which will be contained entirely in $S$. But I am stuck at how to prove that this set is in fact open explicitly. Any help is appreciated.
I tried taking distance of points from the boundary lines of the graph to show that all points are interior, but all I get are useless results which don't help at all.
 A: Notice that
$$S = \{(x_1, x_2 ) \in \mathbb{R}^2 \, : \, x_2 - |x_1| > 0 \} = f^{-1}((0,\infty))$$ where $f \, : \, \mathbb{R}^2 \to \mathbb{R}$ is the continuous function given by $f(x_1, x_2) = x_2 - |x_1|$. Since $S$ is the pullback of an open set under a continuous map, it is open.
A: As you've already noticed, the region is the portion of the plane above the graph of $y=|x|.$ Put another way, it is the region above the graphs of both $y=x$ and $y=-x.$
Taking an arbitrary $\langle x',y'\rangle$ in the region, it's clear that the distance to the line $y=x$ is the length of the segment from $\langle x',y'\rangle$ to the intersection of $y=x$ with the line $y-y'=-(x-x').$ This intersection occurs where $x=y=\frac12(x'+y'),$ so the distance is $$\sqrt{\left(x'-\frac12(x'+y')\right)^2+\left(y'-\frac12(x'+y')\right)^2}=\frac{\sqrt2}2\sqrt{(x'-y')^2}.$$ Similarly, the distance to the line $y=-x$ is the length of the segment from $\langle x',y'\rangle$ to the intersection of $y=-x$ with the line $y-y'=x-x',$ which occurs where $x=\frac12(x'-y')$ and $y=\frac12(y'-x'),$ so the distance is $$\sqrt{\left(x'-\frac12(x'-y')\right)^2+\left(y'-\frac12(y'-x')\right)^2}=\frac{\sqrt2}2\sqrt{(x'+y')^2}.$$
What you need to prove is that both of these are necessarily positive, regardless of which $\langle x',y'\rangle$ we choose from the region. Once that's shown, all we need to do is put $\epsilon>0$ such that $\epsilon$ is less than the minimum of these two distances, at which point the open ball about $\langle x',y'\rangle$ of radius $\epsilon$ lies within the region, concluding the proof.
