Let E be a bounded set in $\mathbb{R}^d.$ Then there exist a closed ball $B(0,K),$ $K>0$ such that $E \subset B(0,K).$ Show that there exist a continuous $f$ on $\mathbb{R}^d$ and $0\leq f(x) \leq 1$ and $f(x)=0$ $\forall x \in E$ and $f(x)=1$ $\forall x \notin B(0,K)$.
I want to use Urysohn’s Theorem to prove this statement but I have a problem that E is only a bounded set and $\mathbb{R}^d - B(0,K)$ is a open set. Please help me to do this proof.
In general it is not true. As a counterexample take $E = B(0,K)$.
However, for each $E$ there exist a minimal $K > 0$ such that $E \subset B(0,K)$ (unless $E = \{0\}$ in which case $K = 0$). Then for each $K' > K$ we can find the desired function without using Urysohn. Simply take $f(x) = 0$ for $\lVert x \rVert \le K$, $f(x) = 1$ for $\lVert x \rVert \ge K'$ and interpolate linearly for $K \le \lVert x \rVert \le K'$.