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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!

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  • $\begingroup$ Is there some relationship between $K$ and $n$ that you aren't stating? Or, have you accidentally used $n$ in several places where you meant $K$? $\endgroup$ – Nick Peterson Nov 25 '15 at 17:41
  • $\begingroup$ Ah you are right, I thought I was making more sense by talking about it in terms of $n$ but I realise now I didn't need to, I will edit it accordingly. $\endgroup$ – Tim Hodgkin Nov 25 '15 at 20:42
  • $\begingroup$ What exactly is your definition of a separator? For instance, for vertex sets $A,B \subseteq V$ with $A \cap B = \varnothing$ you can define a vertex separator for $A$ and $B$ as a subset $S \subseteq V \setminus (A \cup B)$ that separates $A$ and $B$, and such a set might not always exist. If you want to separate a single vertex from the rest of the graph, you must necessarily disconnect it from all its neighbours, so now the problem is trivial... $\endgroup$ – Josse van Dobben de Bruyn Nov 26 '15 at 9:41
  • $\begingroup$ An edge separator is a set of edges that leave at least 2 separate components behind when removed, and a vertex separator is the same with vertices instead of edges. $\endgroup$ – Tim Hodgkin Nov 26 '15 at 11:39
  • $\begingroup$ Okay, so there does not exist a vertex separator for $K_n$, right? Whichever set of vertices is removed, the remaining graph is always connected (and isomorphic to a complete graph on $m$ vertices for some natural number $m \leq n$). $\endgroup$ – Josse van Dobben de Bruyn Nov 27 '15 at 16:33
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Here is a hint for the edge separator: write $K_n = (V,E)$, where $V$ denotes the set of vertices and $E$ the set of edges. Now suppose that the removal of some edge set $F \subseteq E$ leaves the graph disconnected, then we can choose a partition $V = A \cup B$ of the vertices (with $A \cap B = \varnothing$, $A \neq \varnothing$ and $B \neq \varnothing$) so that there are no edges going between $A$ and $B$ in the graph $G \setminus F$. Count the number of edges between $A$ and $B$ in the original graph $K_n$ and so some calculus; this gives a lower bound on $|F|$.

I am currently under the impression that there does not exist a vertex separator. If a vertex separator were to be defined as a vertex set $U \subseteq V$ so that the removal of $U$ leaves the graph disconnected (rather than leaving 2 or more components), then $V$ is the only vertex separator, as the empty graph is disconnected. But this set has cardinality $n$, not $n-1$...

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