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Motivation. For any graph $G$, let $L(G)$ denote its line graph. A graph $G$ with $G\cong L(G)$ is said to be line-graph (LG)-invariant. It turns out that the only finite connected LG-invariant graphs are the cycle graphs. In these graphs, all vertices have degree $2$.

Taking this to the infinite. The only (infinite) connected LG-invariant graph that I found with the property that not all vertices have the same degree is

$G = (\mathbb{N}, E)$ where $E = \big\{\{n,n+1\}:n\in\mathbb{N}\big\}$.

Question. Is there $E\subseteq [\mathbb{N}]^2=\big\{\{x,y\}:x\neq y\in\mathbb{N}\big\}$ such that $G=(\mathbb{N},E)$ is connected, LG-invariant, and there is $N_0\in\mathbb{N}$ such that for all $n\in\mathbb{N}$ with $n\geq N_0$ there is a vertex of $G$ with degree $n$?

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    $\begingroup$ What's an example of an LG-invariant graph with a vertex of degree greater than $2$? $\endgroup$ Commented Feb 5 at 23:55
  • $\begingroup$ Good point @bof, I was convinced that taking some infinite tree would do the job, or the construction used here would yield bigger degrees. But it is well possible that I am wrong. $\endgroup$ Commented Feb 6 at 7:54
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    $\begingroup$ Thanks for the reference. Evidently I read that old question and answer (I left a comment) bit it completely slipped my mind. $\endgroup$ Commented Feb 6 at 9:02
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    $\begingroup$ Doesn’t the Rado graph also provide an example (of infinite degree, as in that question linked by Dominic)? $\endgroup$ Commented Feb 7 at 0:13
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    $\begingroup$ Actually, the Rado graph doesn’t work. The point is that, in any line graph (and any induced subgraph of it) the link of a vertex is covered by two cliques (as an edge in the original graph only has two ends). So line graphs do not contain every finite graphs as induced subgraphs, hence are not isomorphic to the Rado graph. $\endgroup$ Commented Feb 7 at 8:48

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