A compact set inside a bounded open set admits an explicit compactly supported continuous cutoff

If $K=\mathrm{\varnothing }$, the zero function belongs to $C_c({\mathbb{R}}^n)$ and [L3, given, algebra] already satisfies the conclusion. So assume from now on that $K$ is nonempty.

Let $u(x):=d(x,O^c)$. By openness of $O$, one has $u(x)>0$ for every [L1, L2, given, choose, algebra] $x\in K$. The function $u$ is continuous by [L1], so [L2] gives a minimum value $\delta :={\min }_{x\in K}u(x)>0.$

Define $\eta :{\mathbb{R}}^n\to \mathbb{R}$ by [L1, step 1.2, construct] $\eta (x):=\{\begin{array}{ll}0,& u(x)\le \delta \mathrm{/}2,\\ 2u(x)\mathrm{/}\delta -1,& \delta \mathrm{/}2<u(x)<\delta ,\\ 1,& u(x)\ge \delta \mathrm{.}\end{array}$ Because $\eta$ is obtained by composing the continuous function $u$ with a continuous piecewise-linear cutoff on $[0,\mathrm{\infty })$, it is continuous and $0\le \eta \le 1$. Since $u\ge \delta$ on $K$, one has $\eta =1$ on $K$.

The nonzero set of $\eta$ is contained in $\{x:u(x)>\delta \mathrm{/}2\}$, so [L1, L2, L3, step 2.1, algebra] The support of a function on ${\mathbb{R}}^n$ and its compactly supported Riemann integral gives $\mathrm{supp}(\eta )\subseteq \{x:u(x)\ge \delta \mathrm{/}2\}\mathrm{.}$ Every point of the right-hand set lies in $O$, because $u(x)>0$ means $x\mathrm{\notin }{O}^c$. Also $O$ is bounded, so $\bar{O}$ is compact by [L2], and the closed set $\{u\ge \delta \mathrm{/}2\}$ lies in $\bar{O}$. Hence $\mathrm{supp}(\eta )$ is compact and contained in $O$, so $\eta \in C_c({\mathbb{R}}^n)$.