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Question: If $S \subseteq \mathbb{C}$ is open and $L \subset S$ is closed and bounded, show that there exists a $\delta$ such that $|z-w|>\delta$ for all $z \in L$ and $w \notin S$

Attempt: I am considering the set $P := K^C \cap S$ which is open as $K^C$ is, so for every element $a$ of $P$, we can find an $r>0$ such that this is the radius of an open ball centered at $a$ contained in $P$. I now want to set $\delta$ to be the minimum of all such $r$ such that $r$ is the maximum for a particular point $a$. However, I am not sure how to show that this set even has a minimum element, and I feel that this probably isn't even the case. I can imagine a set in which $P$ kind of bottlenecks and the set of all $r$ wouldn't have a minimum, but an infimum of $0$.

user208480
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  • this may be of interest: https://math.stackexchange.com/questions/1952518 – 311411 Mar 21 '21 at 04:34
  • Since $L$ and $S^c$ are disjoint closed sets, the above is implied by the Hausdorff distance being a valid distance i.e. that $d_H(L,S^c) > 0$. I will search for a duplicate in this regard. Look up proof of "Hausdorff distance is a distance" – Sarvesh Ravichandran Iyer Mar 21 '21 at 07:18

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