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.

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

  • $\begingroup$ 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. $\endgroup$ Dec 27 '18 at 16:32
  • $\begingroup$ 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 $. $\endgroup$ Dec 27 '18 at 16:38
  • 1
    $\begingroup$ you are right! I think that's the step I am missing. Thank you! $\endgroup$ Dec 27 '18 at 16:41
  • $\begingroup$ you are welcome! $\endgroup$ Dec 27 '18 at 16:48

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