In a series of lemmas to prove the correctness of a minimum spanning tree algorithm for a finite connected weighted non-directed graph $G$, I come across the statement in the textbook Algorithm Design by John Kleinberg, Eva Tardos:
A cycle and a cutset intersect in an even number of edges
The authors defines a cut of a graph as simply a subset $S \subset G$ of nodes (this may conflict with other definitions of a "cut"). Correspondingly, a cutset is defined as the subset of edges with exactly one endpoint in $S$. The author gives a proof/argument for this statement with a simple picture but I would like to write a more formal argument.
Thoughts: Assume we have a cutset and a cycle whose intersection contains at least one edge. For each edge in the cycle that intersects the cutset, only one node can belong to the cut, so that any edge in the cycle where both nodes are in the cut is not in the intersection.
Consider the next edge in the cycle, either both nodes are in the cutset (an edge not in the intersection), or only one node is in the cutset. In the former case, this cannot continue indefinitely since eventually you would come across an edge from a cutset element to the original node not in the cutset, and so there must eventually be an edge from a cut node to a non-cut node.
Thus for every edge in the intersection entering the cut (an edge from non-cut to cut), there will be a corresponding edge leaving the cut (an edge from cut to non-cut)
Since this is true for each edge in the intersection of the cutset and our cycle, then the overall intersection must have an even cardinality.
Any insights how to clean up this argument appreciated.
A cycle can be viewed as a path starting with a vertex $v$ and finishes in the same vertex $v$. according to your definition of a cut-set as the set of all edges with one endpoint at a subset of vertices $S$, color all the vertices of $S$ blue, color the rest of the vertices red. our vertex $v$ is either blue or red, let's assume for simplicity that it's blue and begin traveling through our cycle. our cycle starts at $v$ with the color blue, it also finishes at $v$ with the color blue. our path changes and only changes color when passing through an edge in the cut-set, and since our path finishes at the same color it started at, it means we must have passed through edges of our cut-set an even number of times.