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Sorry, I couldn't find anything and without proper names it's even harder. It sounds like something reasonably simple to consider and name though.

Given a set of sets $\bf S$, we define some kind of upward closure: $\delta({\bf S})=\{T\subseteq\bigcup {\bf S}\mid T \textit{ maximal such that for any }a,b\in T\textit{ there is }S\in{\bf S}\textit{ with }a,b\in S\}$

A few motivational words: For any $T\in\delta({\bf S})$ there will be some $S\in {\bf S}$ with $T\subseteq S$, if $a,b\in\bigcup{\bf S}$ are never member of the same set $S\in{\bf S}$ they won't be for any $T\in\delta({\bf S})$ either, in a way $\delta({\bf S})$ is an upward closure of $\bf S$ for these properties. And in some way, at least to me, set theoretic topology has very similar operators for filters and filter bases.

I came up with a more elegant definition: $\delta({\bf S})=\{T\subseteq\bigcup {\bf S}\mid b\in\bigcup{\bf S}\setminus T\textit{ iff there is }a\in T\textit{ with no }S\in{\bf S}\textit{ such that }a,b\in S\}$.

Upon consideration I think I'm going to call it a boundary operator, combined with an interior operator $Int$ on $\bf S$ such that $Int({\bf S})=\{T\subseteq 2^{\bigcup{\bf S}}\mid\textit{there is }S\in{\bf S}\textit{ with }T\subseteq S\}$ and a closure operator $cl({\bf S})=Int(\delta({\bf S}))$.

The intuition here is that the boundary are the maximally justified sets considering $\bf S$, the interior are all smaller or equal justified sets considering $\bf S$, the closure is a combination.

  • If ${\bf S}={{0,1},{1},{1,2}}$, would $[{\bf S}]$ be ${{0,1},{1,2}}$? If $\bf S$ equalled ${X\subseteq\Bbb N:|X|=2}$, would $[{\bf S}]$ be $\Bbb N$? Just seeing if I understand the concept. – Akiva Weinberger Nov 12 '15 at 19:03
  • Yes, I think you understand the concept. Just a typo I suppose, in the second case it would be ${{\mathbb N}}$. – pinkwerther Nov 13 '15 at 13:05

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