Prove a property of Minimum Spanning Tree

Question

I need your help proving the following: G=(V,E) is an undirected connected graph with non-negative weights on the edges. Let T be a MST of G and T' be a spanning tree of G (not a minimum one) so it holds that Weight(T') > Weight(T). Prove that the weight of the haviest edge in T' isn't smaller than the weight of heaviest edge in T.

I'm not sure how to approch this, may if we let e(u,v) - heaviest edge on T and e'(u',v') - heaviest edge on T' and then if we look at the cut defined by (u,u') we can use Kruskal algorithem and show that e' isn't chosen to be in T or something in this direction...

Thank you,

Answer 1

Assume the contrary -- there exists a weighted undirected graph with a minimum spanning tree T and a spanning tree T' such that the heaviest edge of T is heavier than the heaviest edge of T', i.e., the heaviest edge of T is heavier than every edge in T'. Consider the cut induced by deleting the heaviest edge h of T. Since T' is connected, some edge in T' crosses this cut. If we add this edge to T - h, we get a spanning tree that is lighter than T, which is a minimum spanning tree. Contradiction.