Prove that $E = \{(x,y): x^2 +4y^2 \le 1\} $ is closed

Question

$$ E = \{(x,y): x^2 +4y^2 \le 1\} $$

How to prove this? I don't know how to set radius of this set

Answer 1

Let $a=(x_0,y_0) \in E^c$, hence

$(*) \quad x_0^2+4y_0^2 >1$.

Now suppose that no $B_r(a) \subset E^c$. Hence, for each $n \in \mathbb N$ there is $(x_n,y_n) \in B_{1/n}(a)$ with $(x_n,y_n) \in E$.

Therefore $x_n \to x_0$ and $y_n \to y_0$. Since $x_n^2+4y_n^2 \le 1$ for all $n$, we get $x_0^2+4y_0^2 \le 1$. But this contradicts $(*)$.

Answer 2

$f(x,y) = x^2 + 4y^2$ is continuous.
$(-\infty,1]$ is closed, so $E = f^{-1}\{(-\infty,1]\}$ is closed.

Answer 3

What you have is an ellipse including it's boundary. There are many ways to go about this question. 1. Show that the complement of E is open. 2. Show that E is the pre-image of a closed set. 3. Show that if you had any converging sequence of points in E, the limit must lie in E.

Answer 4

Here is another easier approach.

Consider a sequence $\{(x_n,y_n)\}_{n\in\mathbb{N}}\subset E$ such that $(x_n,y_n)\rightarrow(p,q)$ as $n\rightarrow\infty$ [Existence of such sequences is always assured by constant sequences]

We now show $(p,q)\in E$.

Consider, $f:E\rightarrow \mathbb{R}$ as $f(x,y)=x^2+4y^2-1$ for all $(x,y)\in E$. Then, clearly $f$ is continuous in $E$ and $f(x_n,y_n)\le 0$ for all $n\in\mathbb{N}$, since $f$ is continuous and the sequence converges to $(p,q)$

$\implies$ $\displaystyle\lim_{n\rightarrow\infty}f(x_n,y_n)\le 0$

$\implies f(p,q)\le 0$

$\implies p^2+4q^2\le 1$ [By definition of $f$].

$\implies (p,q)\in E$. This completes the proof.