User:Stixme/Sandbox

Menger's Theorem
Let $G = (V, E)$ be a graph and $A, B ⊆ V$. Then the minimum number of vertices separating $A$ from $B$ in $G$ is equal to the maximum number of disjoint $A–B$ paths in $G$.

Separation
For $A, B, W \in G(V)$ let $P_i$ denote a path from $A$ to $B$, such as:


 * let $p_s$ and $p_t$ denote the first and the last vertices in $P_i$ respectively.


 * $P_i \cap A = \{p_s\}$ and $\: P_i \cap B = \{p_t\}$ (note that ($P_i \backslash \{p_s,p_t\}) \cap (A\cup B) = \emptyset $)


 * $P_i \cap W \neq \emptyset$

Let $\mathcal{P} = \{P_1, P_2, .., P_n\}$.

If $\mathcal{P}$ contains all possible paths from $A$ to $B$ then we say that set $W$ separates $A$ from $B$.

Let us define $k( G, A, B )$ to be the minimum number of vertices in $W$.

Stronger Statement
Suppose $k \equiv k(G,A,B)$.

Let $\mathcal{P'}, \: \mathcal{P'} \subseteq \mathcal{P}$, and $| \mathcal{P'}\ | = n (\:0 \leq n \leq k − 1\:)$

Then $\exists \mathcal{Q}, \mathcal{Q} = \{Q_1, Q_2, .., Q_{n+1}\}$ such that:
 * if $b \in B$ is one of the endpoints of $P_j$ then $b$ will also be and endpoint of one of the $Q_j$