# Does MST exclude the edge belongs to the cycle of the graph?

For any simple, connected, undirected graph $G(V,E,w)$ with each carrying distinct positive edge weights, is it true that MST will contain all the edges which do not lie in any cycle of $G$?

I think MST can take any edge of irrespective of the fact whether it is an edge of cycle (like I simply use Kruskal's), Also I am unable to see Why the above statement should true. Can anyone tell how the statement is true?

• By the way, all these have nothing to do with whether the edge weights are distinct or not. By the way, an edge that is not in any cycle is called a bridge. So we can say, any MST and in fact, any ST must contain all bridges. – Apass.Jack Aug 17 '18 at 21:56

I think you need to be more precise in your reading of the statement, though it could be better phrased. A different version is "if $e$ is an edge of $G$ and it is not in any cycle in $G$, it must be in any MST of $G$". This doesn't say that edges that are in cycles can't be in the MST, it just says that any edge that isn't in a cycle must be.
It is simply not true that "MST can take any edge of irrespective of the fact whether it is an edge of cycle (like I simply use Kruskal's)". Think of Prim's Algorithm. Every iteration divides the graph into two sets - $Set A$ always forms a single tree. Each step adds to $tree A$ a light edge that connects A to an isolated vertex in $Set B$ (i.e., set of vertices on which no edge of A is incident).
Now imagine a situation where a vertex of $Set B$ is reachable through only one edge i.e., it is not a part of a cycle. To include it in a spanning tree, that one edge becomes necessary. Hence, every edge not belonging to a cycle should be added, else the vertices incident on the other end (belonging to B) of that edge shall never be included into the spanning tree.