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{\bf Let $G$ be a connected graph.}

If $G =$ ~~~~~~~~~~~~ = ~~~~~ , then $\kappa(G) = $

In all other cases:

$\kappa(G)$  =  the size of a minimal vertex cut

~~~~~~~ 	   =  the minimum number of vertices one can
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~~~~~~~~~~  remove such that the resulting subgraph 
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~~~~~~~~~~ is disconnected.

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Thm 2.4: $\kappa(G) \leq \lambda(G) \leq min\{\delta(v) ~|~v \in V(G)\}$


Case 1:  Suppose $G = K_n$, then 
$\kappa(G) = ~~~~~~= \lambda(G)$ 


Case 2:  Let $G$ be a graph such that $ \lambda(G) = k$
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\centerline{Let $E^* = \{e_1, e_2, ..., e_k\}$ be a minimal edge cut of $G$.}

Claim:  $G - E^* = G_1 \cup G_2$ where $G_i$, $i = 1, 2$ are the connected components of $G - E^*$.

Claim:   $e_i = <u_i, v_i>$ where 

\rightline{$u_i \in V(G_1)$ and 
 $v_i \in V(G_2)$ for $i = 1, ..., k$.}




Let $U^* = \{u_1, ..., u_k\} \subset V(G_1)$. ~ Note $|U^*| \leq k$.

Let $V^* = \{v_1, ..., v_k\} \subset V(G_2)$.  ~~~Note $|V^*| \leq k$.






Case 2a:  $\exists u \in V(G_1)$ such that $u \not\in \{u_1, ..., u_k\}$.

Claim:  $u$ is not connected to $v_1$ in $G - U^*$.  

Thus $G - U^*$ is disconnected and hence $U^*$ is a vertex cut for $G$.  Therefore $\kappa(G) \leq k = \lambda(G)$.

Case 2b:  $\exists v \in V(G_2)$ such that $v \not\in \{v_1, ..., v_k\}$.

Claim:  $v$ is not connected to $u_1$ in $G - V^*$.  

Thus $G - V^*$ is disconnected and hence $V^*$ is a vertex cut for $G$.  Therefore $\kappa(G) \leq k = \lambda(G)$.


Case 2c:  $V(G_1) = U^*= \{u_1, ..., u_k\}$ and $V(G_2) = V^*= \{v_1, ..., v_k\} $.


Since $G$ is not a complete graph, 

\centerline{$\exists x, y \in V(G) =  \{u_1, ..., u_k, v_1, ..., v_k\} $}

\centerline{such that $<x, y> ~\not\in~ E(G)$.  }



WLOG assume $x = u_1$.  

Let $N(u_1) = \{u_{i_1}, ..., u_{i_\ell}, v_{j_1}, ..., e_{j_m}\}$  

\rightline{where $u_{i_s} \in U^*$ and  $v_{j_t} \in V^*$ $\forall s, t$.}

Note $x = u_1$ is not connected to $y$ in $G - N(u_1)$.  Thus $N(u_1)$ is a vertex cut for $G$.

Claim $|N(u_1)| =  \delta(u_1) \leq k$.

Define $f:  N(u_1) \rightarrow E^*$ by 

~~~~$f(v_{j_t}) = <u_1, v_{j_t}>$ and  

~~~~$ f(u_{i_s}) = <u_{i_s}, v_p>$ 

\rightline{where $p = min\{j ~|~ v_j \hbox{ is adjacent to } u_{i_s}\}$ }

Note $f$ is a well-defined 1:1 function.

Thus $|N(u_1)| ~\leq~ |E^*|$.  




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$G = (V, E)$ is {\it $k$-connected} if removing any set of $k-1$ vertices in $G$ does not disconnect it and $G \not= $

$K_n$ is ~~~~~~-connected.

$k$ connected implies $k-1$ connected if $k > 1$

%disconnected (more than one component).

connected = 1-connected 



$\kappa(G) = $ connectivity of $G$ = $max\{k ~|~ G ~k-connected \}$
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Example of a 2-edge connected graph:
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Defn 2.13:  A {\bf Harary graph $H_{k, n}$} is a k-connected graph with $n$ vertices and a minimal number of edges.

Note $E(H_{k, n})= \lceil{kn \over 2}  \rceil$ 

\rightline{= the smallest natural number $\geq {kn \over 2}$.}


Thm 2.6:  If $k$ is even, then $G = (V, E)$ is a Harary graph where

 $V = \{0, 2, ...,n-1\}$ and 

$E = \bigcup\limits_{i \in V} \{<i, i + j> ~|~ j = 1, ..., {k \over 2}\}$

%\rightline{$ \cup \bigcup\limits_{i \in V} \{<i, i - j> ~|~ j = 1, ..., {k \over 2}\}$}

where addition is performed modulo $n$.
















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