Distance between the null space of a non-negative function and a set is zero, is the function uniformly lower bounded?

Question

Consider a continuous function $f: \mathbb{R}^n\to\mathbb{R}_{\geq 0}$. Define the set $A=\{x: f(x)=0\}$ which is closed. Assume that $dist(A,B)>0$ where $B$ is some arbitrary set and $dist(A,B)=\inf_{x\in A, y\in B}|x-y|$. Can we show the existence of $\epsilon>0$ such that $f(x)\geq\epsilon$ for all $x\in B$? It seems intuitive but I cannot find a rigorous proof. Thanks for helping.

Answer 1

It is not true. Imagine a continuous function $f: \mathbb{R} \to \mathbb{R}_+$ which is zero when $x \leq 0$ and positive but converges to zero when $x \to \infty$. Then $A = (-\infty, 0]$. For any $a > 0$, let $B = [a, \infty)$. Then $\text{dist}(A, B) = a$ but we can't find positive lower bound $\epsilon$ for $f$ on $B$.

If we restrict $B$ to be compact, then it can be proved.