# graphs – Can an edge that is never a clear edge be part of the minimal spanning tree?

Let $$G = (V, E)$$ to be an undirected graph with arbitrary weighting. Let us define a Cut $$C = (S, V setminus S)$$ of $$G$$ to be a partition of $$G$$ in two sets, $$S$$ and $$V setminus S$$.

For a cut $$C$$, say one edge $$e$$ crosses $$C$$ if and only if an end point of $$e$$ is a summit in $$S$$ and the other is a summit $$V setminus$$. Finally call an edge $$e$$ be a clear edge running through the cup $$C$$ if and only if $$e$$ crosses the cup $$C$$ and has a minimum weight compared to all the other edges that cross the cut $$C$$.

My question is: can an edge $$e$$ it's never a light edge for any cut $$C$$ be part of a tree covering minimum of $$G$$?

I think this is true, simply because we can consider $$G$$ to be a tree then select an edge and make its weight bigger than all the others. Then, as there is only a minimal covering tree, this cutting edge will surely be part of it, and for any cut of $$G$$ it will never be a light edge. Is my reasoning correct?