I am having trouble on this homework question:
Given a K-connected graph $G$ K_n on $n$ vertices where $n \geq 2$, show that the size of the smallest edge separator is the same as the size of the smallest vertex separator, and is size $n-1$.
(the graph K_n is a fully connected graph on n vertices)
Intuitively I can see that to separate a vertex $x$, you must remove all $n-1$ edges attached to $x$, because if even one remains then there will still be a path to $x$from any other vertex through that edge, so $n-1$ edges should be removed. Again for vertex separator if you don't remove all vertices but one, the remaining vertices will all be connected by the property of $G$ being K-connected.
My problem is how to get this across mathematically, one idea I had for the edge separator was to first show that the separator must be of at least size $n-1$ and then show it is at most size $n-1$ thus meaning it must be exactly $n-1$ but again I'm not sure how fully go about this.
Any tips to get me started would be great, thanks!