# Let $\epsilon>0$ and $f\in \mathscr D(\Omega).$ Then Prove that following sets $K_1$ and $K_2$ are compact and disjoint.

Let $$\Omega \subseteq \mathbb R^n$$ be an open set. $$\mathscr D(\Omega)=\{f|f \in C^\infty(\Omega) \wedge \text{supp}(f)\subseteq \Omega \text{ is compact.}\}$$

Let $$\epsilon>0$$ and $$f\in \mathscr D(\Omega).$$ Then Prove that $$K_1=\{ x\in \Omega| f(x)\ge \epsilon\}$$ and $$K_2= \{ x\in \Omega| f(x)\leq -\epsilon\}$$ are compact and disjoint.

My attempt:- I know the definition of $$\text{supp}(f)=\overline{\{ x\in \mathbb R^n: f(x)\neq 0\}}.$$

Disjointness of $$K_1$$ and $$K_2$$

Let $$x\in K_1\cap K_2\implies f(x)\leq -\epsilon \wedge f(x) \ge \epsilon$$ No such exists. If exists, It violates the well definition of $$f.$$ $$K_1=f^{-1}([\epsilon, \infty))$$ and $$K_2=f^{-1}((-\infty, -\epsilon])$$ are closed. ($$\because f$$ is a continuous function.)

I don'tknow how to proceed.

Thank you.

• Something seriously wrong. $K_1$ and $K_2$ are obviously not disjoint. Commented Jul 10 at 8:53
• youtube.com/… Commented Jul 10 at 9:07
• @geetha290krm could you check 19:20 of this video by NPTEL? He says like that. Commented Jul 10 at 9:08
• You made a horrible mistake in the definition of $K_2$. Commented Jul 10 at 9:15
• $K_1$ and $K_2$ are subsets of the support of $f$ (which is compact). Closed subsets of compact sets are compact. Commented Jul 10 at 9:31