In the question I have to decide whether the set $S=\{(x,y)\in\mathbb{R}^2\;|\;x/y\leq 7\}$ is open, closed or neither. I attempted to prove it was closed but it turns out it is neither can someone tell me whats wrong here. So let $A=\{(x,y)\in\mathbb{R}^2\;|\;x/y > 7\}=\{(x,y)\in\mathbb{R}^2\;|\;y>0,\;x -7y > 0\}\cup\{(x,y)\in\mathbb{R}^2\;|\;y<0,\;x -7y < 0\}$
Set $g(x,y)=y$, $ f(x,y)=x-7g$ then g and f are cts functions. Then
$A=(g^{-1}(0,\infty)\cap f^{-1}(0,\infty))\cup(g^{-1}(-\infty,0)\cap f^{-1}(-\infty,0))$
So A open by cty of f and g so $\mathbb{R}^2$ \ $A=S$ is closed.
At the very end of the proof, you said that $\mathbb R^2\setminus A=S$. You overlooked the fact that points where $y=0$ are in neither $A$ nor $S$, because $x/y$ is undefined there.