Talk:Ore Graph is Connected

Limited to $n \ge 3$?
Is it necessary for it to state that the number of vertices is greater than or equal to $3$?

It can in fact be noted that the null graph, the single-vertex graph and the 2-vertex connected graph are all Ore vacuously. The order 2 disconnected graph is not Ore.

So, vacuously, all these graphs can be shown to fulfil the conditions of this theorem. So the condition on $n \ge 3$ is superfluous. Or is there something I am missing? --prime mover (talk) 21:07, 12 June 2016 (UTC)


 * You are right. But then we have to change the proof to explicitly treat the case $n < 3$, because in this case the two components may not be well defined.


 * For $n = 1$ we note that $G$ is trivially connected. For $n = 2$ the proof follows in the same way as it does for $n \ge 3$ except that we understand (intuitively) that the degree of each vertex is zero. No special case needed.


 * Also note that there is only one Ore graph of order $3$, and for such a graph there are no two non-adjacent vertices. The only connected order 3 graph with non-adjacent vertices is not Ore. So if we really want to start with non-vacuously Ore graphs, we would need to start with $n = 4$ in which case the only Ore graph is the $4$ cycle graph. --prime mover (talk) 21:25, 12 June 2016 (UTC)


 * The existence of two non-adjacent vertices is guaranteed by the existence of the two components which we suppose that exist to prove by contraction, so, there is no need to suppose $n \ge 3$. I edited the proof and I think it is better now. -- Hilder.vitor (talk) 22:30, 12 June 2016 (UTC)