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May 8, 2018 at 17:40 comment added Dominic van der Zypen Thanks for your comments! - I understood right away what you meant by "indiscrete" and I often wonder whether there is some kind of functor (or weaker connection) between the category of topological spaces and the category of simple, undirected graphs, such that indiscrete spaces correspond exactly to edgeless graphs (and discrete spaces to complete graphs).
May 8, 2018 at 17:16 comment added Joel David Hamkins I should have said "edgeless" subgraph, since I think "indiscrete" is not the right word for this.
May 8, 2018 at 14:46 comment added Joel David Hamkins An observation: every graph has maximal indiscrete subgraphs -- subgraphs with no induced edges -- and these are vertex transitive. But they are not necessarily maximal with respect to being vertex transitive, and so this doesn't answer the question.
May 8, 2018 at 14:31 comment added Joel David Hamkins Mahdi, if the whole graph is vertex transitive, then it would be maximal.
May 8, 2018 at 14:26 comment added Mahdi - Free Palestine You possibly assume that $S \neq V$ in the precise formulation of the problem.
May 8, 2018 at 14:03 history asked Dominic van der Zypen CC BY-SA 4.0