Let $A$ be a collection of disjoint closed sets such that for every two distinct elements $A_k, A_{k'} \in A$ with $k \neq k'$ we have $$\epsilon = \inf_{x_k \in A_k, x_{k'} \in A_{k'}}|x_k - x_k'| > 0.$$ That is, all pairs of sets are not only disjoint but their elements are at least $\epsilon$ away from each other. Is it true that the union of any collection of such sets is also closed? For $\epsilon = 0$ this is not true (as answered here). However, for $\epsilon > 0$ all Cauchy sequences must eventually lie in only one of the sets. Thus, their limit points must also be in the set. Consequently, the union must contain all limit points which means it is closed. Is my reasoning correct?
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Is there a single $\epsilon$ lower bound for all pairs, or each pair its own independent lower bound? – Chad K Jan 31 '24 at 19:19
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Good question. Let's say that there is a single $\epsilon$ lower bound for all pairs. – Jakob Jan 31 '24 at 19:22
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For a single common lower bound your reasoning is correct. – Chad K Jan 31 '24 at 19:26