Equivalence of Definitions of Matroid Rank Axioms/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:
\((\text R 1)\) | $:$ | \(\ds \map \rho \O = 0 \) | |||||||
\((\text R 2)\) | $:$ | \(\ds \forall X \in \powerset S \land y \in S:\) | \(\ds \map \rho X \le \map \rho {X \cup \set y} \le \map \rho X + 1 \) | ||||||
\((\text R 3)\) | $:$ | \(\ds \forall X \in \powerset S \land y, z \in S:\) | \(\ds \map \rho {X \cup \set y} = \map \rho {X \cup \set z} = \map \rho X \implies \map \rho {X \cup \set y \cup \set z} = \map \rho X \) |
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 $A \cup \set {b_j} \in \mathscr I$ |
- $\map \rho A = \map \rho {A \cup \set {b_j}}$
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:
- $\map \rho {A \cup \set{b_1, b_2, \dots, b_{i-1}, y } } = \map \rho {A \cup \set{b_1, b_2, \dots, b_{i-1} } }$
- $\map \rho {A \cup \set{b_1, b_2, \dots, b_{i-1}, b_i, y } } = \map \rho {A \cup \set{b_1, b_2, \dots, b_{i-1} } } = \map \rho A$
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:
- $\map \rho {A \cup B} = \map \rho A$
$\blacksquare$