I've been thinking on this problem for half a day, and I'm stumped. I've tried constructing $C$ directly, but that doesn't seem to be a good idea. In the case $K$ turns out to be a compact $\textbf{interval}$, I've seen answers here; I am curious as to whether it works for general compact sets.
I've also considered the collection $\mathcal{C}$ of all compact subsets of $\mathbb{R}$ and tried showing that at least one of $C\in\mathcal{C}$ satisfy $f(C)=K$; I think something like Zorn's lemma should be required, and I'm not making progress.
Thanks in advance.
- The problem isn't from a book, I'm trying to use it as a lemma for my work, so I am not sure as to whether the statement is true or not.
Best Answer
Let $a=\inf K$ and $b=\sup K$. By Let $f : I \to \mathbb R$ be continuous. For any compact interval $J \subseteq f(I), \exists$ a compact interval $K \subseteq I$ with $f(K)=J.$ we see that there is an interval $[c,d]$ with $f([c,d])=[a,b])$. Let $C =[c,d] \cap f^{-1} (K)$. Check that this $C$ serves the purpose.