User:Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Rank Axioms/Condition 1 Implies Condition 3/Lemma 3

Theorem
Let $S$ be a finite set.

Let $\rho : \powerset S \to \Z$ be a mapping from the power set of $S$ to the integers.

Let $\rho$ satisfy the rank axioms:

Let:
 * $A \subseteq S : \map \rho A = \card A$

Let:
 * $B \subseteq S : \forall b \in B \setminus A : \map \rho {A \cup \set b} \ne \card{A \cup \set b}$

Then:
 * $\map \rho {A \cup B} = \map \rho A$

Case 1 : $B \setminus A = \O$
Let $B \setminus A = \O$.

From Set Difference with Superset is Empty Set:
 * $B \subseteq A$

From Union with Superset is Superset:
 * $A \cup B = A$

It follows that:
 * $\map \rho {A \cup B} = \map \rho A$

Case 2 : $B \setminus A \ne \O$
Let:
 * $B \setminus A = \set{b_1, b_2, \dots, b_k}$

For each $i \in \closedint 1 k$, let:
 * $B_i = \begin{cases}

A & : i = 1 \\ A \cup \set{x_1, x_2, \dots, x_{i-1}} & : i \in \closedint 2 k \end{cases}$


 * $\mathscr B_i = \set{B_i \cup \set y : y \in B \setminus B_i}$

We note that:

We prove by induction:
 * $\forall i \in \closedint 1 k : \forall Y \in \mathscr B_i : \map \rho Y = \map \rho A$

For all $i \in \closedint 1 k$, let $\map P i$ be the proposition:
 * $\forall Y \in \mathscr B_i : \map \rho Y = \map \rho A$

Basis for the Induction
We have:
 * $\mathscr B_1 = \set{A \cup \set y : y \in \set{b_1, b_2, \dots, b_k}}$

$\map P 1$ is the case:
 * $\forall j \in \closedint 1 k : \map \rho {A \cup \set{b_j}} = \map \rho A$

Let $j \in \closedint 1 k$.

We have:

By rank axiom $(\text R 2)$:
 * $\map \rho A = \map \rho {A \cup \set {b_j}}$

This establishes our base case.

Induction Hypothesis
Now we need to show that, if $\map P i$ is true, where $i \ge 1$, then it logically follows that $\map P {i + 1}$ is true.

So this is our induction hypothesis:
 * $\forall Y \in \mathscr B_i : \map \rho Y = \map \rho A$

Then we need to show:
 * $\forall Y \in \mathscr B_{i+1} : \map \rho Y = \map \rho A$

Induction Step
This is our induction step.

Let $Y \in \mathscr B_{i+1}$.

Then:
 * $Y = A \cup \set{b_1, b_2, \dots, b_i, y}$

where $y \in {b_{i+1}, \dots, b_k}$

We note that:
 * $A \cup \set{b_1, b_2, \dots, b_{i-1}, b_i} \in \mathscr B_i$
 * $A \cup \set{b_1, b_2, \dots, b_{i-1}, y} \in \mathscr B_i$

We have:

Hence:
 * $\map \rho {A \cup \set{b_1, b_2, \dots, b_{i-1}, b_i } } = \map \rho {A \cup \set{b_1, b_2, \dots, b_{i-1} } }$

and
 * $\map \rho {A \cup \set{b_1, b_2, \dots, b_{i-1} } } = \map \rho A$

Similarly:
 * $\map \rho {A \cup \set{b_1, b_2, \dots, b_{i-1}, y } } = \map \rho {A \cup \set{b_1, b_2, \dots, b_{i-1} } }$

By Rank axiom $(\text R 3)$:
 * $\map \rho {A \cup \set{b_1, b_2, \dots, b_{i-1}, b_i, y } } = \map \rho {A \cup \set{b_1, b_2, \dots, b_{i-1} } } = \map \rho A$

It follows that:
 * $\forall Y \in \mathscr B_{i+1} : \map \rho Y = \map \rho A$

This establishes our induction step.

Hence by induction:
 * $\forall i \in \closedint 1 k : \forall Y \in \mathscr B_i : \map \rho Y = \map \rho A$

We have:

In either case:
 * $\map \rho {A \cup B} = \map \rho A$