# How to show that $\{(x, y) \in \mathbb{R}^2\mid\cos(x^2) + x^3 - 4 7y > e^x - y^2\}$ is open or closed?

Show that the set of points $$(x, y) \in \mathbb{R}^2$$ such that $$\cos(x^2) + x^3 - 4 7y > e^x - y^2$$ is an open subset of $$\mathbb{R}^2$$.

I was reading the fact that if $$(X,d)$$ and $$(Y,d)$$ are metric spaces and $$f:(X,d) \to \rightarrow (Y,d)$$ is continuous and if $$v$$ be a open set (or closed set) in $$Y$$ then $$f^{-1}(v)$$ is open (closed) in $$X$$. I don't really under stand how to do the applications for this lemma. So far I have proved using this theorem that the identity function and the constant function are continuous. I want help to understand this problem and for similar problems I want to know what I am looking for here.what are the steps to show that a given set is closed or open using this lemma.

$$f(x,y) = \cos(x^2) + x^3 - 47y - (e^x - y^2)$$ is continuous. And the set you're looking for is $$f^{-1}[(0, \infty)]$$, i.e. the inverse image of an open subset of $$\mathbb R$$ under a continuous map. Hence this set is open.

• Sir i understand the part that f is continuous. I am asking what are the steps to find f^-1[(0, infinity)]. – Sinchan Bhattacharjee May 25 at 8:21
• You don't need to find $f^{-1}((0,\infty))$ to show that it is open. – Minus One-Twelfth May 25 at 8:27
• @SinchanBhattacharjee You just have to convince yourself that $f^{-1}[(0, \infty)] = \{(x,y) \in \mathbb R^2 \mid \cos(x^2) + x^3 - 4 7y > e^x - y^2\}$. – mathcounterexamples.net May 25 at 8:55

$$f(x,y)=\cos (x^{2})+x^{3}-47y-e^{x}+y^{2}$$ defines a continuous function and the given set is $$f^{-1} (0,\infty)$$.

The other answers show how to do it for this specific example, but it’s helpful to see it as a general principle: A strict inequality between continuous functions always defines an open set.

Precisely, given continuous functions $$l, r : X \to \mathbb{R}$$ on any metric space $$X$$ (more generally, any topological space), the set $$\{ x \in X\ |\ l(x) < r(x) \}$$ is open. This is because the function $$f : X \to \mathbb{R}$$ given by $$f(x) := r(x) - l(x)$$ is continuous, and $$\{ x \in X\ |\ l(x) < r(x) \} = \{ x \in X\ |\ f(x) > 0 \} = f^{-1}(0,\infty)$$.

Exercise: with a similar argument, show that a non-strict inequality between continuous functions always defines a closed set, i.e. that any set of the form $$\{ x \in X\ |\ l(x) \leq r(x) \}$$ (for continuous functions $$l, r : X \to \mathbb{R}$$) is closed.