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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.

Comments

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.

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  • $\begingroup$ 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. $\endgroup$
    – José Carlos Santos
    Oct 16, 2020 at 13:53

1 Answer 1

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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.

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  • $\begingroup$ Ah yes I found that while I was looking online but I believe that comes in the next section so I don't think we're suppose to use the theorem that if the inverse set is closed, so is the set. $\endgroup$
    – John Miller
    Oct 16, 2020 at 13:57
  • $\begingroup$ 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. $\endgroup$
    – John Hopfensperger
    Oct 16, 2020 at 14:00
  • $\begingroup$ 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 $\mathbb{R}^2$? $\endgroup$
    – John Miller
    Oct 16, 2020 at 14:03
  • $\begingroup$ 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 $U\subset Y$. $\endgroup$
    – John Hopfensperger
    Oct 16, 2020 at 14:06
  • $\begingroup$ 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. $\endgroup$
    – John Hopfensperger
    Oct 16, 2020 at 14:09

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