# Show that the area enclosed by an ellipse $E$ centred at 0 is an open set.

Show the set $$E = \{(x,y) : a\cdot x^2 + b\cdot y^2 \leq R^2 \}$$ with $$a,b,R \in \mathbb{R}$$ is a closed set.

My attempt

I figured showing that $$E' = \{(x,y) : a\cdot x^2 + b\cdot y^2 > R^2 \}$$ is open would be simpler as we can define $$\partial E'= \{(x,y): a\cdot x^2 + b\cdot y^2 = R^2\}$$. Then given $$(x_0, y_0) \in E'$$ let $$r = \min\{|(x,y) - (x_0, y_0)| : (x,y)\in \partial E'\}$$ we need to show $$r>0$$. By contradiction if is not greater than zero then there exists a sequence $$(x_n, y_n) \to (x_0, y_0)$$ with $$(x_n, y_n) \in \partial E'$$ for every $$n \in \mathbb{N}$$ but since $$\partial E'$$ is closed (the function $$f(x,y) = a\cdot x^2 + b\cdot y^2$$ is clearly continuous for all pairs $$(a,b)$$ by algebra of continuity) thus $$(x_0, y_0) \in \partial E'$$ so we have a contradiction thus $$r>0$$ and the set $$E'$$ is open, so its complement $$E$$ is closed.

The one part I'm not sure about is the contradiction of $$r>0$$. I believe we're just trying to show $$r \neq 0$$ since $$|.|\geq 0$$ and since $$(x_0, y_0)$$ must lie inside $$\partial E'$$ for $$r=0$$ thus $$(x_0, y_0)\in \partial E'$$ which is a contradiction since we said $$(x_0, y_0) \in E'$$ and $$\partial E' = \bar{E'} \cap \bar{E'}'$$ so $$\partial E' \cap E' = \emptyset$$ so $$(x_0, y_0)$$ can't be in both sets. Should I of made this clearer in my proof or is the sufficient. I'm just starting my second year course in multivariable calculus so I'm not use to the proof style yet.

• You cannot simply define $\partial E'$. The boundary of the set $E'$ is $\{(x,y)\mid ax^2+by^2=R^2\}$, but proving that is probably harder than proving directly that $E$ is a closed set. Oct 16, 2020 at 13:53

I think you can express $$E$$ in the form $$f^{-1}([0, R^2])$$ where $$f: \mathbb{R}^2 \to \mathbb{R}$$ is a continuous function.
• You will basically be using this fact no matter what. To give a "direct" proof, you will pick a point $(x,y)$ such that $ax^2 + by^2 > R$ and show that you can perturb $x$ and $y$ some small amount while maintaining the inequality. Essentially, you will be showing $f$ is continuous. Oct 16, 2020 at 14:00
• One thing to note it says that these are equivalent: i) $f:U \to \mathbb{R}^k$ is continuous at all points in $U$. ii) For all open subsets $V$ of $\mathbb{R}^k$, $f^{-1}(V)$ is relatively open to $U$. So in this case $E$ is relatively open to $\mathbb{R}^2$ so it's open in $\mathbb{R}^2$? Oct 16, 2020 at 14:03
• For most mathematicians, the definition of a continuous function $g: X\to Y$ is a map such that $g^{-1}(U)$ is open for each open $U\subset Y$. Oct 16, 2020 at 14:06
• The word "relative" doesn't seem useful here. If I have a space $X$ and a subset $Y \subset X$, then the "relative topology" on $Y$ consists of all sets of the form $U \cap X$ where $U$ is open in $X$. But this doesn't seem relevant to our problem. Oct 16, 2020 at 14:09