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Fake Proof: Finding mistakes in a “proof” of Menger's theorem

1

$begingroup$

I am working on the following exercise:

Find all mistakes the following ‘proof’ of the set version of Menger’s
theorem. What statement(s) would be necessary to prove in order to complete the
proof?

Let S be a smallest $A-B$ separator. We say that a component $C$ of $G-S$ meets $A$ if $C$ contains a vertex of $A$. Denote by $G_A$ the graph that $G$ induces on the union of $S$ and all vertex sets of components of $G-S$ that meet $A$. Define $G_B$ analogously.

By the choice of $S$ and induction, $G_A$ contains $|S|$ disjoint $A-S$ paths, while $G_B$ contains $|S|$ disjoint $S-B$ paths. Joining these paths yields the desired set of $|S|$ disjoint $A-B$ paths.

In my opinion there is a huge gap in the part "By the choice of $S$ and induction, $G_A$ contains $|S|$ disjoint $A-S$ paths, while $G_B$ contains $|S|$ disjoint $S-B$ paths." I do not really understand what is happening here.

Do you think that this is a sufficient answer for this exercise?

graph-theory fake-proofs

share|cite|improve this question

asked Mar 18 at 11:19

3nondatur3nondatur

404111

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is the induction hypothesis?
  $endgroup$
  – hbm
  Mar 18 at 15:39
  

  
  
  
  

add a comment | 

1

$begingroup$

I am working on the following exercise:

Find all mistakes the following ‘proof’ of the set version of Menger’s
theorem. What statement(s) would be necessary to prove in order to complete the
proof?

Let S be a smallest $A-B$ separator. We say that a component $C$ of $G-S$ meets $A$ if $C$ contains a vertex of $A$. Denote by $G_A$ the graph that $G$ induces on the union of $S$ and all vertex sets of components of $G-S$ that meet $A$. Define $G_B$ analogously.

By the choice of $S$ and induction, $G_A$ contains $|S|$ disjoint $A-S$ paths, while $G_B$ contains $|S|$ disjoint $S-B$ paths. Joining these paths yields the desired set of $|S|$ disjoint $A-B$ paths.

In my opinion there is a huge gap in the part "By the choice of $S$ and induction, $G_A$ contains $|S|$ disjoint $A-S$ paths, while $G_B$ contains $|S|$ disjoint $S-B$ paths." I do not really understand what is happening here.

Do you think that this is a sufficient answer for this exercise?

graph-theory fake-proofs

share|cite|improve this question

asked Mar 18 at 11:19

3nondatur3nondatur

404111

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is the induction hypothesis?
  $endgroup$
  – hbm
  Mar 18 at 15:39
  

  
  
  
  

add a comment | 

$begingroup$

I am working on the following exercise:

Find all mistakes the following ‘proof’ of the set version of Menger’s
theorem. What statement(s) would be necessary to prove in order to complete the
proof?

Let S be a smallest $A-B$ separator. We say that a component $C$ of $G-S$ meets $A$ if $C$ contains a vertex of $A$. Denote by $G_A$ the graph that $G$ induces on the union of $S$ and all vertex sets of components of $G-S$ that meet $A$. Define $G_B$ analogously.

By the choice of $S$ and induction, $G_A$ contains $|S|$ disjoint $A-S$ paths, while $G_B$ contains $|S|$ disjoint $S-B$ paths. Joining these paths yields the desired set of $|S|$ disjoint $A-B$ paths.

In my opinion there is a huge gap in the part "By the choice of $S$ and induction, $G_A$ contains $|S|$ disjoint $A-S$ paths, while $G_B$ contains $|S|$ disjoint $S-B$ paths." I do not really understand what is happening here.

Do you think that this is a sufficient answer for this exercise?

graph-theory fake-proofs

share|cite|improve this question

asked Mar 18 at 11:19

3nondatur3nondatur

404111

$endgroup$

share|cite|improve this question

asked Mar 18 at 11:19

3nondatur3nondatur

404111

share|cite|improve this question

asked Mar 18 at 11:19

3nondatur3nondatur

404111

asked Mar 18 at 11:19

3nondatur3nondatur

404111

asked Mar 18 at 11:19

3nondatur3nondatur

404111