Do inequalities that define certain regions imply if the region is closed/open?

Question

I'm asked to prove whether or not A = {(x, y) | y > f(x)} is an open subset of \mathbb{R}^{2}} when f is continuous. My approach was to start by assuming A is closed (with the intent of proving by contrapositive), so it's complement A = {(x, y) | y <= f(x)} is open, with the hopes of showing that y <= f(x) implies some sort of boundedness in all cases, and that this is enough to show that A is an open subset. Am I on the right track?

Answer 1

A proof by contrapositive would start with negating the statement "$A$ is open," which would be "$A$ is not open" rather than "$A$ is closed."

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You can prove the statement directly. Suppose $(x,y) \in A$. We construct a neighborhood of $(x,y)$ that lies entirely in $A$. Let $\epsilon = y - f(x) > 0$. There exists some small interval $(x-\delta, x+\delta)$ such that $f(u) < f(x) + \epsilon/2$ for all $u \in (x-\delta, x+\delta)$. Then $A$ contains the rectangle $\{(u,v) : u \in (x-\delta, x+\delta), v \in (f(x) + \epsilon/2, f(x) + 3\epsilon/2)\}$ which in turn contains the original point $(x,y)$.