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How can it be proved that the set of all $(x,y)$ such that $3x^2 + 2y^2<6$ is an open set?

I tried to prove directly the aforementioned statement. Without success I tried to prove that the image of a linear mapping applied in a open set is also an open set. I do not know if this last part is true, but I could not prove it. Can someone help me? Thank you.

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2 Answers 2

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$f(x,y)=3x^2+2y^2$ is a continuous function from $R^2$ to $R$ and $(-\infty,6)$ is an open subset of $R$, so $f^{-1}(-\infty,6)$ is an open subset of $R^2$.

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Write $g(x,y)=3x^2+2y^2-6$, and note that $g^{-1}((-\infty,0))$ is the inverse of an open subset, thus is open since $g$ is continuous.

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