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.

  • $\begingroup$ Something seriously wrong. $K_1$ and $K_2$ are obviously not disjoint. $\endgroup$ Commented Jul 10 at 8:53
  • $\begingroup$ youtube.com/… $\endgroup$
    – Unknown x
    Commented Jul 10 at 9:07
  • $\begingroup$ @geetha290krm could you check 19:20 of this video by NPTEL? He says like that. $\endgroup$
    – Unknown x
    Commented Jul 10 at 9:08
  • $\begingroup$ You made a horrible mistake in the definition of $K_2$. $\endgroup$ Commented Jul 10 at 9:15
  • 2
    $\begingroup$ $K_1$ and $K_2$ are subsets of the support of $f$ (which is compact). Closed subsets of compact sets are compact. $\endgroup$ Commented Jul 10 at 9:31


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