Why the set of points satisfying (2x-x²-y²)*(x²+y²-x)>0 is the same of the set with condition ((2x-x²-y²)>o and (x²+y²-x)>0)?

Question

Why the set of points on $\Bbb R^{2}$ satisfying $(2x-x²-y²)(x²+y²-x)>0$ is the same of the set of all elements on $\Bbb R^{2}$ with condition $((2x-x²-y²)>0 $ and $(x²+y²-x)>0)$, i think that the first set includes elements that is not in the last one, for example the elements satisfying $((2x-x²-y²)<0 $ and $(x²+y²-x)<0)$

Also how can i proof that these set must be open?

Thank you.

Answer 1

This is because the set of solutions of the other system: $$\begin{cases}2x-x^2-y^2<0,\\x^2+y^2-x<0\end{cases}$$ is empty.

This an open set because it's the intersection of the inverse images of the open set $]0,+\infty[$ by two polynomial functions, which are continuous.

Answer 2

Suppose $x^2+y^2-x < 0$. Then we necessarily have $x > 0$, and hence $2x-x^2-y^2 = x-(x^2+y^2-x) > x + 0 > 0$, and we've showed there exist no solutions for that system.