Graph theory (Edge Connectivity)

Question

Suppose $\lambda = k \gt 0$. Show that there are sets of vertices $U$ and $V$ that partition the vertices of $G$, and such that there are exactly $k$ edges with one endpoint in $U$ and one endpoint in $V$. I literally don't know how to start this problem.... Help me, please......

Answer 1

If $G$ is $k$-edge connected, then you can select $k$ edges as a separating set. That is to say removing your $k$ edges breaks $G$ into exactly $2$ components, say $U$ and $V$. This partitions the vertices of $G$ between the two connected components you create when the $k$ edges are removed. Every edge in a separating set necessarily has one end in $U$ and the other end in $V$. If not, an edge could be added back in and still leave $U$ and $V$ separated, which would contradict the assumption that $k$ edges are a minimum edge separating set.