Let's say we have a graph $T$, which is also a spanning tree of a graph $G$, $T$ is connected, now let's say the diameter of $T$ is $x$.
Now, what would be the diameter of :
$$ (T \cup T^c) + (T \cup T^c) $$
where $T^c$ is the complement of $T$, and we assume that $T$, $T^c$ are the different copies of $T$.
What I know is since $T$ and $T^c$ are disjoint, the union of them will be a disconnected graph, but I'm not clear how the join operation denoted by $+$ sign here change the final resulting graph. Is it still disconnected? (which I hope not)
My idea was that the union of $T$ and $T^c$ would be a complete graph, but it seems like it's wrong if I plot a sample graph. Hence, I'm looking for help. Thank you for any support in advance.
The union of $T$ and its complement $T^{c}$ is a complete graph $K_n$, where $n$ is the number of nodes in $T$. The diameter of a complete graph is ofcourse 1, since all vertices are connected to each other by distance 1. Hence, since the join of graphs $G+H$ connects every node of $G$ to each one of $H$, we obtain that the problem's join graph $$(T\cup T^c) + (T\cup T^c) = K_n + K_n = K_{2n}$$ is also complete, so its diameter is 1.