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How does exactly the induction go in the proof number one? What is the induction hypothesis there and what is the induction step?

By the induction hypothesis, $G/e$ contains an $A–B$ separator $Y$ of fewer than $k$ vertices.

Whole theorem and proof is here, page 67.

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up vote 5 down vote accepted

As it says at the beginning of the proof, the induction is on $\|G\|$, the number of edges of $G$. The induction hypothesis is that Menger’s theorem is true for all graphs with fewer edges than $G$.

The proof then proceeds by contradiction. Suppose that $k$ is the minimum number of vertices separating $A$ and $B$ in $G$, but, contrary to Menger’s theorem, $G$ does not have $k$ disjoint $A$-$B$ paths. We’re assuming at this point that $\|G\|\ge 1$, so $G$ has an edge $e=xy$. Let $v_e$ be the contraction vertex in $G/e$, and set

$$A'=\begin{cases} A,&\text{if }A\cap\{x,y\}=\varnothing\\ \big(A\setminus\{x,y\}\big)\cup\{v_e\},&\text{if }A\cap\{x,y\}\ne\varnothing \end{cases}$$


$$B\,'=\begin{cases} B,&\text{if }B\cap\{x,y\}=\varnothing\\ \big(B\setminus\{x,y\}\big)\cup\{v_e\},&\text{if }B\cap\{x,y\}\ne\varnothing\;. \end{cases}$$

An $A'$-$B\,'$ path in $G/e$ that does not contain $v_e$ is an $A$-$B$ path in $G$, and an $A'$-$B\,'$ path in $G/e$ that does contain $v_e$ induces an $A$-$B$ path in $G$ in an obvious way. Moreover, it’s clear that disjoint $A'$-$B\,'$ paths in $G/e$ correspond to disjoint $A$-$B$ paths in $G$, so $G/e$ must have fewer than $k$ disjoint $A'$-$B\,'$ paths. Since $\|G/e\|=\|G\|-1<\|G\|$, the conclusion of Menger’s theorem hold for $G/e$ by the induction hypothesis, so $G/e$ has an $A'$-$B\,'$ separator $Y$ with fewer than $k$ vertices. This is the first of two points in the induction step at which the induction hypothesis is used; the argument continues as follows.

If the contraction vertex $v_e$ were not in $Y$, then $Y$ would be an $A$-$B$ separator in $G$, so $v_e\in Y$. Let $X=\big(Y\setminus\{v_e\}\big)\cup\{x,y\}$; then $X$ is an $A$-$B$ separator in $G$. By hypothesis $|X|\ge k$; on the other hand, $|X|=|Y|+1$, and $|Y|<k$, so we must have $|X|=k$. We now have an $A$-$B$ separator with exactly $k$ vertices.

Now suppose that $S$ is an $A$-$X$ separator in $G-e$. Every $A$-$B$ path in $G$ must pass through $X$ and therefore through $S$, so $S$ is an $A$-$B$ separator in $G$; thus, $|S|\ge k$. $\|G-e\|=\|G\|-1<\|G\|$, so the conclusion of Menger’s theorem holds for $G-e$, which therefore has $k$ disjoint $A$-$X$ paths. This is the second point in the induction step at which the induction hypothesis is used. Similarly, $G-e$ has $k$ disjoint $X$-$B$ paths. If one of these $A$-$X$ paths met one of these $X$-$B$ paths outside of $X$, we could construct from them an $A$-$B$ path in $G$ that did not go through $X$, which is impossible, so the $A$-$X$ paths are disjoint from the $X$-$B$ paths outside of $X$. Being disjoint, the $A$-$X$ paths must terminate in distinct vertices of $X$. Similarly, each of the $X$-$B$ paths begins at a distinct vertex of $X$. Thus, each of the $k$ $A$-$X$ paths $\pi$ can be paired up with the $X$-$B$ path beginning at the terminal vertex of $\pi$ to make an $A$-$B$ path, and the $k$ $A$-$B$ paths formed in this way are disjoint. That completes the proof.

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Thanks for answer! What if $x \in A$ and $y \in B$? Should it be mentioned as a spesial case, where Menger theorem still stands or does this proof include it? – Ohto Nordberg Dec 20 '12 at 15:17
@Ohto: You’re welcome! It’s not a problem: in that case $v_e$ belongs to both $A'$ and $B\,'$, but it’s still one of the vertices in $Y$, and the proof goes through unchanged. – Brian M. Scott Dec 20 '12 at 16:45
Ok, I see :). Now I fail to see the contradiction. The structure of this proof keeps eluding me ... What things become contradictory and where does it take this proof? I also have hard time understanding, why the graph G-e is considered. Is it because contracting "$/$" eliminates vertices and deleting "$-$" eliminates edges? – Ohto Nordberg Dec 20 '12 at 20:34
@Ohto: There are a couple of sub-arguments that use proof by contradiction, but overall it’s not a proof by contradiction: it’s a proof by induction on the number of edges, and the induction step shows how the desired $A$-$B$ paths. It uses $G/e$ to show that $G$ has an $A$-$B$ separator $X$ with exactly $k$ vertices, and then it uses $G-e$ to show that there must be $k$ disjoint paths from $A$ to $X$ and $k$ from $X$ to $B$ that can be ‘sewn together’ to make $k$ disjoint paths from $A$ to $B$. – Brian M. Scott Dec 21 '12 at 0:34

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