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For a fixed set $X$ and a finite collection $E_1,E_2,\ldots,E_k\subseteq X$, define the binary relation adjacency as follows: $E_i,E_j$ are adjacent if their intersection is nonempty. We term the transitive closure of this relation by transitive adjacency and define the adjacent union by $$ \tilde\cup(E_1,\ldots,E_k):= \begin{cases} \bigcup_{i=1}^k E_i, & \text{the $(E_i)$ are transitively adjacent} \\ \emptyset, & \text{else} . \end{cases} $$

Are there standard terms for transitive adjacency and adjacent union?

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    $\begingroup$ As you define it, adjacency is not an equivalence relation, since it is not transitive. And if you apply the transitive closure, then every two non-empty sets $A$ and $B$ become equivalent, because $A$ is adjacent to $A \cup B$, and $A \cup B$ is adjacent to $B$... $\endgroup$
    – Ivan Izmestiev
    Commented Sep 22, 2016 at 13:48
  • $\begingroup$ Good points, will edit! $\endgroup$
    – Aryeh Kontorovich
    Commented Sep 22, 2016 at 14:03
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    $\begingroup$ I see. A collection of subsets is sometimes called a hypergraph, but I don't know if there are already terms for what you are defining. $\endgroup$
    – Ivan Izmestiev
    Commented Sep 22, 2016 at 14:25

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You can think in terms of intersection graphs. The transitive adjacency tells you when two vertices are adjacent. The adjacency union is then empty if and only if two of $E_1,\ldots,E_k$ are in distinct connected components.

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  • $\begingroup$ Thanks! Since asking the question, it's looking like I won't be needing this concept after all -- at least not for now :) $\endgroup$
    – Aryeh Kontorovich
    Commented Oct 21, 2016 at 11:29
  • $\begingroup$ I guess transitive adjacency tells you when two vertices are in the same connected component. $\endgroup$
    – Bjørn Kjos-Hanssen
    Commented Sep 16, 2017 at 6:17
  • $\begingroup$ Exactly. That's another way of stating it. $\endgroup$
    – Vinicius dos Santos
    Commented Sep 16, 2017 at 13:53
  • $\begingroup$ But adjacent means connected by 1 edge of the graph, right? $\endgroup$
    – Bjørn Kjos-Hanssen
    Commented Sep 17, 2017 at 5:46
  • $\begingroup$ In the traditional graph interpretation, yes. $\endgroup$
    – Vinicius dos Santos
    Commented Sep 17, 2017 at 11:11
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In philosophy, this would be called family resemblance -- if $E_i\cap E_j\ne\emptyset$ and $E_j\cap E_k\ne\emptyset$ then $E_i$ and $E_k$ have a family resemblance.

That is, perhaps I have no common feature with my second cousin, but we both have common features with our common great-grandparent. Similarly, there may be no single thing that is common to all "games".

The idea that every-day concepts such as game and number exhibit family resemblance comes from Wittgenstein's notes from 1930, so perhaps it is prior to the notion of intersection graphs as in @ViniciusdosSantos' answer, which seems to only go back to 1945.

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