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How to show prove that the set $\left\{(x,y)\in\mathbb R^{2} : \frac{x^2}{4}+\frac{y^2}{9}<1\right\}$ is a open set?

my attempt:

Graphically it being a interior of ellipse, without boundary shows that it is open but how to prove it? Can we take the equation of ellipse to be a function and say that its domain is open if its range is?

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  • $\begingroup$ Choose any $(x,y)$ in your set. You need to show that there exists a neighbourhood of $(x,y)$ that is also within the set; in other words, find an $r$ for which the circle centred at $(x,y)$ with radius $r$ is contained within the ellipse. The value of $r$ will depend on $(x,y)$. (Side note: what is $R$? Do you mean $(x,y) \in \mathbb R^2$?) $\endgroup$ – Théophile Aug 13 '15 at 14:56
  • $\begingroup$ Can you perhaps come up with a continuous function $f$ that has this set as a pre-image of an open set? $\endgroup$ – anakhro Aug 13 '15 at 14:57
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It is the pre-image of the open set $(-\infty,1)$ under the continuous mapping $(x,y)\mapsto \frac{x^2}4+\frac{y^2}9$.

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