Element is Member of Base iff Not Loop
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Theorem
Let $M = \struct{S, \mathscr I}$ be a matroid.
Let $\mathscr B$ denote the set of all bases of $M$.
Let $x \in S$.
Then:
- $\exists B \in \mathscr B: x \in B$ if and only if $x$ is not a loop
Proof
Necessary Condition
Let $B \in \mathscr B$ such that $x \in B$.
From Singleton of Element is Subset:
- $\set x \subseteq B$
By definition of a base:
- $B \in \mathscr I$
From matroid axiom $(\text I 2)$:
- $\set x \in \mathscr I$
Then $\set x$ is not a dependent subset by definition.
It follows that $x$ is not a loop by definition.
$\Box$
Sufficient Condition
Let $x$ not be a loop.
By definition of a loop:
- $x$ is not a dependent subset
By definition of a dependent subset:
- $x \in \mathscr I$
From Independent Subset is Contained in Base:
- $\exists B \in \mathscr B: \set x \subseteq B$
By definition of a subset:
- $x \in B$
$\blacksquare$
Sources
- 1976: Dominic Welsh: Matroid Theory ... (previous) ... (next) Chapter $1.$ $\S 4.$ Loops and parallel elements