3
$\begingroup$

Is there a class of graphs whose edge sets are equivalence relations?

This class seems similar to comparability graphs (edge set form a partial order) but require the edge set to be symmetric rather than anti-symmetric.

  • 0
    To be reflexive, a vertex has to have an edge to itself. Depending on your definition of edges in a graph, this might be impossible. – 2017-01-16
  • 0
    Yes, the class of graphs I'm looking for require all vertices to have an edge to itself. I define an edge to be a pair of vertices. In extension, a set of edges (i.e., the edge set) is a binary relation on the set of vertices. – 2017-01-16
  • 0
    The definition of graphs, that I was told forbid edges $(u,u)$, that's why I was asking. Thanks for claryfing. In that case: What about complete Graphs? Since a equivalence asks basically every element in a class to be related to every other element, that's the way to go and has no relation to the "outside". – 2017-01-16
  • 0
    In a complete graph there is an edge between every pair of distinct vertices. Hence a complete graph where the edge set is also reflexive would be a special case of the class I'm looking for. In particular note that the class I'm looking for do not require the graph to be connected. – 2017-01-16
  • 0
    Inside every equivalence class, you must a complete graph to fulfill the requirements for an equivalence relation. If you put in another complete graph (unconnected to the first) you have a second equivalence class. – 2017-01-16
  • 0
    If you define a complete graph to have edges between every pair of vertices then yes, the edge set is an equivalence relation with one class. The disjoint union of N such complete graphs is a complete graph where the edge set is an equivalence relation with N classes. Note that the definition of complete graph I gave require each pair of vertices to be distinct though. – 2017-01-16

0 Answers 0