Prove that $\{x\in \Bbb{R^n} :\;f(x)\leq k\}$ is compact

Question

Let $f:\Bbb{R^n}\to \Bbb{R}$ be continuous such that $$\lim_\limits{||x||\to\infty}f(x)=+\infty.$$

I want to show that $\{x\in \Bbb{R^n}:\; f(x)\leq k\}$ is compact.

I'm thinking that it suffices to show that the set is closed and bounded as a subset of $\Bbb{R^n}.$

I'm stuck at this point. Please, can anyone help me out? Thanks for your time and help!

Answer 1

We can proof that exists $m \in \mathbb{R}^n$, exists $M \in \mathbb{N}$ such that for all $x \in \mathbb{R}^n$ and $||x|| > M$ implies $f(x) \geq f(m)$. Supose that is false, ie, for all $M \in \mathbb{N}$ exists $x \in \mathbb{R}^n$ and $||x|| > M$ implies $f(x) < f(m)$. Then for all $M \in \mathbb{N}$ exists $||x|| > M$ implies $f(x) < f(m)$ means $f$ is bounded from above by $f(m)$ but $f$ is unbounded from above. It is a contradiction.
$f$ is bounded from below, because $f$ is a coersive function, and all coersive function has minimum value.
Let be $k \in \mathbb{R}$ and $B = \{ x \in \mathbb{R}^n : f(x) \leq k \}$ such that $B \neq \varnothing$. Let be $C = \{ x \in \mathbb{R}^n : ||x|| \leq k \}$. $C$ is compact, and $B \cap C$ is compact, because $f$ is continuous. Therefore $B$ is compact.