 Solution

Assume the graph is connected so that a minimum spanning tree exists, and let $T$ be the output of Kruskal's algorithm. With distinct weights there is a strict total edge ordering. Suppose another minimum spanning tree $T^{\prime }\ne T$ existed. Choose $e\in T\setminus T^{\prime }$ and increase only $w(e)$ by a positive $\epsilon$ so small that the strict ordering of all edge weights remains unchanged. For a finite graph such an $\epsilon$ exists, using less than the gap to the next larger weight; if $e$ is largest, any sufficiently small positive increase works.

Running Kruskal's algorithm again examines exactly the same order and makes exactly the same component decisions. It therefore still returns $T$, which must remain optimal for the perturbed weights. But $T$ has become more expensive by $\epsilon$, while $T^{\prime }$ is unchanged. They had equal original weight, so now $T^{\prime }$ is cheaper, a contradiction. This proves that distinct weights imply a unique minimum spanning tree.

The converse is false. A triangle with edge weights $1,1,2$ has a unique minimum spanning tree, consisting of its two weight-one edges. Its total is $2$, while either other spanning tree has total $3$. Thus equal edge weights permit, but do not force, multiple optimum trees. The same perturbation proof applies componentwise to a minimum spanning forest.