How to show that $\Omega_{c}=\left\{x \in \mathbb R^{n}: V(x) \leq c\right\}$ is compact?

Question

Consider an ODE system $$\dot{x}=f(x),$$ having a candidate Lypunov function, which satisfies $V(x)\geq0$, $V(0)=0$, and $\dot{V}(x)\leq0$.

How to show that the set $$\Omega_{c}=\left\{x \in \mathbb R^{n}: V(x) \leq c\right\}$$ is compact?

Since $V(x)$ is a radially unbounded function, such that $\Omega_{c}$ must have an upper bound, $0$ will be the lower bound of the set.

What about the closed property? We know this set must contain the lower boundary $0$, how do we know whether this set contains an upper boundary?

Answer 1

Let $c > 0$. We have to show that $\Omega_c$ is closed and bounded.

$\newcommand{\R}{\mathbb{R}}$

1. Since $V$ is differentiable, $V:\R^n \to \R$ is continuous. Therefore, $\Omega_c$ is closed.

2. From the definition of radial unboundedness $$ \forall c > 0: \exists r > 0: \forall x \in \R^n : (||x|| > r \Rightarrow V(x) > c) $$ where the contrapositive of the conditional is $$ V(x) \le c \Rightarrow ||x|| \le r $$ Therefore, $\Omega_c$ is bounded.