Discussion was held on the solution to the planar graph exercise assigned last class.

Advice on proofs:

 - Typically when using induction to prove some property of graphs, it may
not be best to proceed by adding a vertex or edge.  The reason is that there quite likely
will be some countably infinite ways in which to do this.

- One approach might be to remove a single vertex or edge, show that the
property holds, and then proceed by adding it back to the graph in the same place.

- For clarity and organization it is a good idea to separate sub-proofs out
into lemma's.  This not only helps to organize, but it will also draw attention
to important points along the way to proving/disproving the central hypothesis.
 

Dr. Sherman's Solution to the planar graph problem:

Proof (by induction on the number of edges)
    Let S = { m  Î Natural #'s: For all connected, planar graphs G, with m edges,
                  V_G + R_G - E_G = 2 } ( j will be assigned to represent this property ).

    Note: Be specific in defining S

Basis Must show:  0 Î S
    The only connected graph with 0 edges has only 1 vertex, in which case V_G = 1,
    R_G =  1, and E_G = 0.  Therefore, 0 Î S.

Inductive Step Let m Î Natural #'s, and assume m Î S; we must show that m + 1 Î S.

    Let G be any connected planar graph with m + 1 edges; we must show that
V_G + R_G - E_G = 2.

    Note that G is either a tree, or G has a cycle C.  This step indicates that the proof has
two cases.

    Case 1:  Assume G is a tree then ... (the solution follows directly from the definition of a tree)

    Case 2:  Assume G has some cycle C.  Let e be any edge in C.  Let G' be the
subgraph of G formed by removing e.  Since m Î S, G' has property j.  Since V_G = V_G' ,
R_G = R_G' + 1, and E_G = E_G' + 1 it follows that G has property j , and thus
m + 1 Î S.

\ Therefore S Î Natural #'s
 

Exercise:

    For any sets A, B, and X prove the following statements are equivalent:

1. A Í A È B
2. (X - A) Ç (X - B) = Æ
3. X - A Í B

Form a cycle of implications between the statements.

No complete solution was arrived at in class although there was a general agreement
that statement 1 looked as though it may be in error.