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

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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:

\((\text R 1)\)   $:$   \(\displaystyle \map \rho \O = 0 \)             
\((\text R 2)\)   $:$     \(\displaystyle \forall X \in \powerset S \land y \in S:\) \(\displaystyle \map \rho X \le \map \rho {X \cup \set y} \le \map \rho X + 1 \)             
\((\text R 3)\)   $:$     \(\displaystyle \forall X \in \powerset S \land y, z \in S:\) \(\displaystyle \map \rho {X \cup \set y} = \map \rho {X \cup \set z} = \map \rho X \implies \map \rho {X \cup \set y \cup \set z} = \map \rho X \)             



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:

$\map \rho {A \cup B} = \map \rho A$


Proof

Lemma 1

$\forall A, B \subseteq S: A \subseteq B \implies \map \rho A \le \map \rho B$


Lemma 2

$\forall A \subseteq S: \map \rho A \le \card A$


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:

$\map \rho {A \cup B} = \map \rho A$

$\Box$


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:

\(\ds \mathscr B_k\) \(=\) \(\ds \set{B_k \cup \set{b_k} }\)
\(\ds \) \(=\) \(\ds \set{A \cup \set{b_1, b_2, \dots, b_{k-1} } \cup \set{b_k} }\)
\(\ds \) \(=\) \(\ds \set{A \cup \set{b_1, b_2, \dots, b_k} }\)
\(\ds \) \(=\) \(\ds \set{A \cup B \setminus B}\)
\(\ds \) \(=\) \(\ds \set{A \cup B}\) Set Difference Union Second Set is Union

$\Box$


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:

\(\ds \map \rho {A \cup \set {b_j} }\) \(<\) \(\ds \card {A \cup \set {b_j} }\) As $\map \rho {A \cup \set {b_j} }$ and from Lemma 2
\(\ds \) \(=\) \(\ds \card A + 1\)
\(\ds \) \(=\) \(\ds \map \rho A + 1\) As $A \cup \set {b_j} \in \mathscr I$

By rank axiom $(\text R 2)$:

$\map \rho A = \map \rho {A \cup \set {b_j}}$

This establishes our base case.

$\Box$


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:

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

$\Box$


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:

\(\ds \map \rho {A \cup \set{b_1, b_2, \dots, b_{i-1} } }\) \(\le\) \(\ds \map \rho {A \cup \set{b_1, b_2, \dots, b_{i-1}, b_i } }\) Rank axiom $(\text R 2)$
\(\ds \) \(=\) \(\ds \map \rho A\) Induction Hypothesis
\(\ds \) \(\le\) \(\ds \map \rho {A \cup \set{b_1, b_2, \dots, b_{i-1} } }\) Lemma 1


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.

$\Box$


Hence by induction:

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


We have:

\(\ds \map \rho A\) \(=\) \(\ds \map \rho {A \cup \set{b_1, b_2, \dots, b_k} }\) As $A \cup \set{b_1, b_2, \dots, b_k} \in \mathscr B_k$
\(\ds \) \(=\) \(\ds \map \rho {A \cup B \setminus A}\)
\(\ds \) \(=\) \(\ds \map \rho {A \cup B}\) Set Difference Union Second Set is Union

$\Box$

In either case:

$\map \rho {A \cup B} = \map \rho A$

$\blacksquare$