Definition
Let $S$ be a finite set.
Let $\mathscr B$ be a non-empty set of subsets of $S$.
$\mathscr B$ is said to satisfy the base axiom if and only if:
| \((\text B 1)\)
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$:$
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\(\ds \forall B_1, B_2 \in \mathscr B:\)
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\(\ds x \in B_1 \setminus B_2 \implies \exists y \in B_2 \setminus B_1 : \paren {B_1 \setminus \set x} \cup \set y \in \mathscr B \)
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$\mathscr B$ is said to satisfy the base axiom if and only if:
| \((\text B 2)\)
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$:$
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\(\ds \forall B_1, B_2 \in \mathscr B:\)
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\(\ds x \in B_1 \setminus B_2 \implies \exists y \in B_2 \setminus B_1 : \paren {B_1 \cup \set y} \setminus \set x \in \mathscr B \)
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$\mathscr B$ is said to satisfy the base axiom if and only if:
| \((\text B 3)\)
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$:$
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\(\ds \forall B_1, B_2 \in \mathscr B:\)
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\(\ds \exists \text{ a bijection } \pi : B_1 \setminus B_2 \to B_2 \setminus B_1 : \forall x \in B_1 \setminus B_2 : \paren {B_1 \setminus \set x } \cup \set {\map \pi x} \in \mathscr B \)
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$\mathscr B$ is said to satisfy the base axiom if and only if:
| \((\text B 4)\)
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$:$
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\(\ds \forall B_1, B_2 \in \mathscr B:\)
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\(\ds x \in B_1 \setminus B_2 \implies \exists y \in B_2 \setminus B_1 : \paren {B_1 \setminus \set x} \cup \set y, \paren {B_2 \setminus \set y} \cup \set x \in \mathscr B \)
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$\mathscr B$ is said to satisfy the base axiom if and only if:
| \((\text B 5)\)
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$:$
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\(\ds \forall B_1, B_2 \in \mathscr B:\)
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\(\ds x \in B_1 \setminus B_2 \implies \exists y \in B_2 \setminus B_1 : \paren {B_2 \setminus \set y} \cup \set x \in \mathscr B \)
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$\mathscr B$ is said to satisfy the base axiom if and only if:
| \((\text B 6)\)
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$:$
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\(\ds \forall B_1, B_2 \in \mathscr B:\)
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\(\ds x \in B_1 \setminus B_2 \implies \exists y \in B_2 \setminus B_1 : \paren {B_2 \cup \set x} \setminus \set y \in \mathscr B \)
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$\mathscr B$ is said to satisfy the base axiom if and only if:
| \((\text B 7)\)
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$:$
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\(\ds \forall B_1, B_2 \in \mathscr B:\)
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\(\ds \exists \text{ a bijection } \pi : B_1 \setminus B_2 \to B_2 \setminus B_1 : \forall x \in B_1 \setminus B_2 : \paren {B_2 \setminus \set {\map \pi x} } \cup \set x \in \mathscr B \)
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See also