Consider an undirected graph $G = (V_G, E_G), |V| = n$ . Let $T = (V_T, E_T)$ be a spanning tree of $G$, and let $T^\prime = (V_{T^\prime}, E_{T^\prime})$ be the complement of $T$ such that e $\in$ $E_{T^\prime}$ $\iff$ e $\notin$ $E_T$. Is it always true that $E_{T}$ $\cup$ $E_{T^\prime} = K_n$ ?
I observed that this is indeed the case for G = $K_2, K_3, K_4, K_5$ (trying to solve a problem involving complete graphs), but I wanted to make sure it held in the general case.
This is trivially the case. Clearly if the union contains all edges in the graph and all edges in the complement, then it is a complete graph because it contains all edges.