Final Answer:
There is only one minimum weight spanning tree in a connected weighted graph G with unique edge weights.
Step-by-step explanation:
In a connected weighted graph where each edge has a unique weight, we can establish the uniqueness of the minimum weight spanning tree (MST) through a proof by contradiction. Suppose there are two distinct MSTs, T1 and T2, in graph G. Since both trees are minimal, removing any edge e from either tree should disconnect the graph, creating two connected components. Let's assume e belongs to T1.
Now, consider the edge with the smallest weight, w, in the symmetric difference (XOR) of T1 and T2. Without loss of generality, let's assume w is in T1 but not in T2. By adding w to T2, we create a cycle. Within this cycle, there exists an edge, e', that is also in T1. Since e' and w have different weights, we can replace e' with w in T1, forming a new spanning tree with a smaller weight.
This contradicts the assumption that T1 is an MST, as we've found a spanning tree with a smaller weight. Therefore, the initial assumption of having two distinct MSTs is false, establishing the uniqueness of the minimum weight spanning tree in a connected weighted graph with unique edge weights.
In summary, the contradiction arises from assuming two distinct MSTs, leading to the conclusion that there can only be one minimum weight spanning tree in a connected weighted graph G with unique edge weights.