how can I show $\{(x, y) : x^{2} > y\}$ is an open set in $\mathbb{R}^{2}$? I have not seen a proof of this kind, so I was wondering if someone can help me with this problem as an example. The book I'm reading doesn't help much.
I thought that maybe I can just say that $\partial S = \{(x, y) : x^{2} = y\}$ is the boundary and $\partial S \cap S = \emptyset$. But then I thought that maybe it isn't enough to state the boundary, and I need to prove that the set you are claiming is the boundary.
Any help is appreciated
The map$$\begin{array}{rccc}f\colon&\mathbb{R}^2&\longrightarrow&\mathbb R\\&(x,y)&\mapsto&x^2-y\end{array}$$is continuous, the set $(0,\infty)$ is an open subset of $\mathbb R$ and your set is $f^{-1}\bigl((0,\infty)\bigr)$