# Compactness and Enlargement

Consider a bounded continuous function that satisfies $$|f(x)|\leq\epsilon$$ on a compact set K. I am asked to prove that there is a $$\delta$$ enlargement of K, $$K^\delta$$ such that $$|f(x)|\leq2\epsilon ,\forall x\in K^\delta$$ $$K^\delta=\{x|d(x,K)<\delta\}$$ for some metric.

The hint says uses the compactness of K. I was trying to use the finite covering property of K but the argument does not go through when I try to select a $$\delta$$. I would like to know if there is some alternative way to prove the above claim.

• What is the domain and range of $f$? Dec 27 '18 at 16:29
• there is no specification of that. x is in a general metric space D and f is a bounded and continuous function. Dec 27 '18 at 16:35

Hint: for each $$x\in K$$ there is a $$\delta_x>0$$ such that $$y\in B(x,\delta_x)\Rightarrow f(y)\in B(f(x),\epsilon).$$ Then, $$K\subseteq \bigcup_{x\in K}B(x,\delta_x)$$ and $$\mathscr A=\{B(x,\delta_x):x\in K\}$$ is a cover of $$K$$. Reduce to a finite subcover and unravel the definitions.
• Then my question is that suppose we have picked the finite cover and found the minimum radius $\delta_{x^*}$ for example. How can we make sure that K^{\delta_{x^*}} is what we want? There may be points that are included in the finite subcover but their $\delta_{x^*}$ balls does not satisfy the desired criterion. Dec 27 '18 at 16:32
• because if $x\in K$ then $y\in B(x,\frac{1}{2}\delta_x)\Rightarrow d(y,K)\le d(y,x)<\delta_x\Rightarrow d(f(y),f(x))<2\epsilon$. Dec 27 '18 at 16:38