Determine whether the following set is compact: $\{(x,y)\} \in \mathbb{R^2} : 2x^2 - y^2 \leq 1\}$

Question

Determine whether the following set is compact: $\{(x,y) \in \mathbb{R^2} : 2x^2 - y^2 \leq 1\}$

The question is asking to just determine, but I would like to know how would you SHOW that this set is compact?

Defn: A set $A$ is compact $\Leftrightarrow$ the set is closed and bounded.

Defn: A set is closed if it contains all of its limit points

Defn: A set is bounded if $\exists$ R such that the set $A$ is contained in the $B_{R}(0)$

So I am able to reason out that this set is compact. The set is bounded because there does exist an $R$ in this case $R = 1$ where this set is contained within $B_{1}(0) \in \mathbb{R^2}$.

As for closed I would have to somehow show that my set contains all of its limit points. To do that I would have to know what all of the limit points are, but that is impossible. Perhaps I could use contradiction and say something along the lines that if a limit point was not in the set it would violate the conditions of the given set. Specifically $$2x^2 - y^2 \leq 1$$

This is how I would think out the problem, but I know this is not formal, so how would I formalize this into a proper proof? What am I missing in my though pattern?

Answer 1

This set is unbounded: for each $x>1$, take $y=\sqrt{2x^2-1}$. Then $(x,y)\in A$, but since $x$ can be arbitrarily large… So, $A$ is not compact.

Answer 2

Lets call your set $H = \{(x,y) \in \mathbb{R^2} : 2x^2 - y^2 \leq 1\}$, because it contains a hyperbola.

Concerning unboundedness:

With $u = \sqrt{2}x-y$ and $v = \sqrt{2}x+y$ you get: $$2x^2-y^2 = (\sqrt{2}x-y)(\sqrt{2}x+y) = uv = 1$$ So, it's unbounded and, hence, not compact.

Concerning closedness:

Let $(x_0,y_0)= \lim_{n\rightarrow\infty}(x_n,y_n)$ with $(x_n,y_n) \in H$ for all $n \in \mathbb{N}$. You just need to show that $(x_0,y_0) \in H$. But this is easy to show as $$(x_0,y_0)= \lim_{n\rightarrow\infty}(x_n,y_n) \Leftrightarrow x_0= \lim_{n\rightarrow\infty}x_n \mbox{ and } y_0= \lim_{n\rightarrow\infty}y_n$$ It follows immediately

$$2x_0^2 - y_0^2 = \lim_{n\rightarrow\infty} (2x_n^2 - y_n^2) \leq 1$$