Showing Component of Superlevel Set is Compact

Question

Let $f \colon \mathbb{R}^2 \to \mathbb{R}$ be defined by $f(x, y) = \frac{x^2}{\left(x^2 + y^2 + 1\right)^3}$. The superlevel set $$D = \left\{(x,y) \in \mathbb{R}^2 \colon f(x,y) \geq \frac{1}{10}\right\}$$ looks like this

and consists of two connected components. I know the function is continuous (and bounded), so I can show that $D$ is compact. My question is, how do I show a single one of the components is compact?

Answer 1

Every connected component of a compact metric space is compact. Proof: Connected components are closed, and a closed subset of a compact space is compact.

Answer 2

Restrict $f$ to an open set, which separates the two component of the superlevel set that means it consists the one component, and disjoint from the other. The restricted function does what you need. :)

Answer 3

Hint: Note that $|f|\leq 1$ (a better bound may exist, but this one is obvious). For $c>0$, write $$D_c=\{(x,y)\in\mathbb R^2 :f(x,y) \geq c\} = f^{-1}([c,1])$$ Since $(0,y)\notin D_c$, and $D_c$ is compact, we can find a "tube" $(-\epsilon,\epsilon)\times \mathbb R$ which doesn't meet $D_c$. Can you see that $D_+$ and $D_-$ are both compact now, where $$D_+ = \underbrace{D_c}_{\textrm{compact}} \cap \underbrace{([\epsilon,\infty)\times\mathbb R)}_{\textrm{closed}}$$ $$D_- = \underbrace{D_c}_{\textrm{compact}} \cap \underbrace{((-\infty,-\epsilon]\times\mathbb R)}_{\textrm{closed}}$$ and $D_{\pm}$ are the components in which you are interested?