Continuous map on a compact set Let $f$ be a continuous map on $\mathbb{R}^d$. We denote $A=\min\limits_{x\in K}f(x).$
I want to prove that there exist $\epsilon_1>0$ such that 
$$f(x)\le \epsilon_1\Rightarrow f(x)\ge \frac{\epsilon_0}{2}$$
Please help me to do so. Thanks
 A: Since $f$ is continuous and $\frac{\epsilon_0}2>0$, we know that for each $p \in K$ the set $f^{-1}\left(\{y\in \mathbb{R}:|y-f(p)|<\frac{\epsilon_0}2\}\right)$ is open in $\mathbb{R}^d$. We also know that $\underset{p \in K}\bigcup f^{-1}\left(\{y\in \mathbb{R}:|y-f(p)|<\frac{\epsilon_0}2\}\right)$ contains $K$. Since the collection of open sets $\big\{ f^{-1}\left(\{y\in \mathbb{R}:|y-f(p)|<\frac{\epsilon_0}2\}\right) \big\}_{p \in K}$ covers $K$, there exists $\epsilon_1>0$ so that for every $p \in K$ we have $B_d(\,p, \epsilon_1) \subseteq f^{-1}\left(\{y\in \mathbb{R}:|y-f(q)|<\frac{\epsilon_0}2\}\right)$ for some $q \in K$ (Lebesgue's Number Lemma). Thus if $d(x,K)<\epsilon_1$, then $|f(x)-f(q)|<\frac{\epsilon_0}2$ for some $q \in K$. And since $f(q)\geq \epsilon_0$, we therefore have $f(x)>f(q)-\frac{\epsilon_0}2\geq \frac{\epsilon_0}2$.
Invoking Lebesgue's Number Lemma allows us to get at the fact that $f$ is uniformly continuous on $K$.
A: Hint: Use the following three facts (prove them!):


*

*For points outside of $K$, we have $d(x,K) = d(x,\partial K)$.

*The boundary $\partial K$ is again compact.

*$f$ is continuous, use the $\epsilon$-$\delta$ formulation of continuity.

