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Aug 9, 2021 at 12:03 comment added Farmer S For infinite graphs, the statement is equivalent to transitivity. (If $G$ is infinite and transitive, and $x\in V$, let $C_x$ be the transitively connected component $\{y\in V\bigm|xEy\}$, and for such a component $C$, let $A_C$ be a subset of $V$ of size $\mathrm{card}(V)$, with the $A_C$'s pairwise disjoint. Then define $\psi(x)=A_{C_x}\backslash\{x\}$, and note this works, using $\kappa=\mathrm{card}(V)$, or using $\kappa=3$, since for $x\neq y$, we have $xEy$ iff $C_x=C_y$ iff $A_{C_x}=A_{C_y}$ iff $\psi(x)\Delta\psi(y)=\{x,y\}$ iff $\psi(x)\Delta\psi(y)$ has card $<\kappa$.)
Aug 9, 2021 at 9:11 vote accept Dominic van der Zypen
Aug 9, 2021 at 0:08 answer added Mikhail Tikhomirov timeline score: 2
Aug 8, 2021 at 14:17 comment added Dominic van der Zypen Good point @FarmerS - thanks for noticing! - I would be delighted in a counterexample for any graph where no $\kappa$, finite or infinite, with the property stated in the question exists
Aug 7, 2021 at 20:29 comment added Farmer S Sorry, if $\kappa$ is infinite, that is.
Aug 7, 2021 at 20:21 comment added Farmer S Doesn‘t the condition imply the graph is transitive?
Aug 7, 2021 at 20:07 history asked Dominic van der Zypen CC BY-SA 4.0