Minimal edge cut Suppose that $C$ is a minimal edge cut of a graph $G=(V,E)$ is it possible that the removal of $C$ can split $G$ into three components? I ask this because i'm reading a proof which states that it's possible however it seems rather counter intuative as it seems any such edge edge cut shouldn't be minimal?
One thing i tried is looking at the following graph vertices $a,b$ and $c$ with an edge from $a$ to $c$ and from $b$ to $c$ then the two edges form an edge cut which split the graph into three components (in this case three independent vertices) however it is not a minimal edge cut.
 A: It sounds like you're right !  
Suppose that $C$ is a minimal edge cut that yields at least 3 components.
Call $G'$ the graph obtained after removing $C$ from $G$.
Put back any edge $e \in C$ in $G'$.  The two endpoints of $e$ belong to at most 2 of the components, so adding $e$ can only reduce the number of components of $G'$ by one.  In other words, $G' + e$ is still disconnected, which implies that $C$ wasn't minimal after all.
A: As I understand it, a minimal cut set is one where no subset is also a cut set. If that's the case then I don't see how you can have three components connected by a minimal set. Here's my reasoning.
First, connect the three components with two edges. Then you can delete either edge and get a pair of components.
Now, connect them with three or more (N) edges. Now, you can choose a subset of (N - 1) edges such that one of the components is disconnected.
I tried drawing various convoluted graphs with chiral C3 symmetry, but none quite worked. This was the best I could do:

where the red edges are my attempt at a minimal cut set. You can see that you could delete four of them and disconnect one of the squares.
