Let $S = \{(x,y)\in \mathbb C: y > 3x+2\}$. Show that $S$ is an open set.
I can imagine what it looks like; a shaded region above a line. I also imagine that we must choose $ε$ (the radius around any $z_0 \in S$) to be less than the distance between $z_0$ and the point on the line at which it is perpendicular. If we let $ε$ to be the distance between $z_0$ and the point on the line which $z_0$ is perpendicular, we would have an open disk around $z_0$ contained in $S$. I can easily Google the formula for the desired distance, but I don't know if this is the best way to approach the problem (and I feel like if that were on a test, I wouldn't know how to derive the equation or it would take too long -- that, and I would have to somehow prove that the distance is minimal between a point and line when it is perpendicular). Is there another way?
Yes: Write $S=f^{-1}(U)$ where $U=(0,+\infty)$ and $f:\mathbb R^2\to\mathbb R$ is defined by $f(x,y)=y-3x-2$. Then $f$ is $______$ and $U$ is $____$ hence $S$ is open.