A proof that if f is the heaviest edge in weight from all the other edges in the circle which it is a part of, then f will not participate in any MST

I cant proof that if f is the heaviest edge in weight from all the other edges in the circle which it is a part of, then f will not participate in any Minimum Spanning Tree. please help.

• This might be considered as wrong. For example, consider a triangle with the same weight on all three edges. Commented Aug 28, 2021 at 0:40
• Does this answer your question? What edges are not in any MST? Commented Aug 28, 2021 at 0:42

Lets assume that it did participate in an MST $$T$$.
Now, removing $$f$$ will split $$T$$ into two connected components - $$T_1,T_2$$. In particular, both vertices of $$f$$ need to be in different connected components. Lets call the vertices $$v_1$$ and $$v_k$$, and w.l.o.g. assume $$v_1\in T_1$$ and $$v_k\in T_2$$.
Lets name all vertices in the cycle by $$v_1,v_2,\dots,v_k$$ such that there is an edge between $$v_i$$ and $$v_{i+1}$$ for all $$i$$ (and of course, $$v_1,v_k$$ are the vertices of $$f$$).
Since $$v_1\in T_1$$ and $$v_k\in T_2$$, and also $$v_i\in T_1\lor v_i\in T_2$$ then there must be some index $$i_0 with $$v_{i_0}\in T_1$$ and $$v_{i_0+1}\in T_2$$. We know that $$(v_{i_0} , v_{i_0 + 1})$$ is an edge in $$G$$, and its weight is lower than that of $$f$$ - and in addition it will connect the two components! Therefore, by replacing $$f$$ with this edge we get another spanning tree with a lower weight, which is a contradiction to the assumption we started with an MST!