# Proving heaviest edge in mst is not heavier than heaviest edge of any possible spanning tree?

How to prove that the heaviest edge in mst is not heavier than heaviest edge of any possible spanning tree? by heaviest edge of any Spanning tree i mean considering any possible Spanning tree, we have to prove that the heaviest edge in mst is not heavier than heaviest edge of that spanning tree.

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– D.W.
Nov 23, 2023 at 8:45
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– D.W.
Nov 23, 2023 at 8:46

Let $$e^*$$ be the heaviest edge in an MST $$T$$, and let $$A$$ and $$B$$ be the vertices of the two connected components of $$T - e^*$$.
Now, by the cut property, $$e^*$$ is a cheapest edge with one endpoint in $$A$$ and one endpoint in $$B$$.
Since any spanning tree $$T'$$ must have an edge $$e$$ with one endpoint in $$A$$ and one in $$B$$, this means that $$T'$$ has an edge of cost at least the cost of $$e^*$$.