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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.

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$f(x,y) = \cos(x^2) + x^3 - 47y - (e^x - y^2)$ 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.

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  • $\begingroup$ Sir i understand the part that f is continuous. I am asking what are the steps to find f^-1[(0, infinity)]. $\endgroup$ – Sinchan Bhattacharjee May 25 at 8:21
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    $\begingroup$ You don't need to find $f^{-1}((0,\infty))$ to show that it is open. $\endgroup$ – Minus One-Twelfth May 25 at 8:27
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    $\begingroup$ @SinchanBhattacharjee You just have to convince yourself that $f^{-1}[(0, \infty)] = \{(x,y) \in \mathbb R^2 \mid \cos(x^2) + x^3 - 4 7y > e^x - y^2\}$. $\endgroup$ – mathcounterexamples.net May 25 at 8:55
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$f(x,y)=\cos (x^{2})+x^{3}-47y-e^{x}+y^{2}$ defines a continuous function and the given set is $f^{-1} (0,\infty)$.

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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.

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