Proving $k(G) \leq k'(G) $ where $k(G)$ is the minumum vertex cut and $k'(G)$ is minimum edge cut.

Question

I want to prove $k(G) \leq k'(G) $ where $k(G)$ is the minumum vertex cut and $k'(G)$ is minimum edge cut. Can I argue as follows:

If $k'(G) $ is any (take minimum for instance) edge cut then form take one vertex end from each of the edges in $k'(G)$. This gives vertex cut as removing these vertex will remove all the edges from $k'(G)$ and so $k(G) \leq k'(G)$.

In books like West I find much more work put in and therefore am afraid if the above argument has any loopholes.

Answer 1

Your argument is a good start, but it does not take into account that the vertices you remove might compose one entire connected component that is left by your edge cut. For example, look at this graph:

        *----*
       /      \
      *--*--*--*
       \      /
        *----*

The two edges connecting the outer cycle to the inner two vertices form a cut set, but following your argument, you could remove the two inner vertices, which is not a vertex cut.