Richard PennerApr 19

Take a set, 𝑥. Is it finite or infinite? Well what definition do you use?

① ∃𝑦 ∈ ω₀ 𝑥 ≈ 𝑦 ; There is a natural number, 𝑦, such that you may may map 𝑦 one-to-one onto the set 𝑥, enumerating each of its members. So 𝑥 is finite for the same reason { 1, 2, 3 } is finite.

② ¬ ∃𝑧 ∈ (On ∖ ω₀) 𝑥 ≈ 𝑧 ; There is no such infinite ordinal, 𝑧, such that you may map 𝑧 one-to-one onto the set 𝑥. So 𝑥 is finite because ω₀, the smallest infinite ordinal, cannot be mapped 1-to-1 into it.

Do ① and ② say the same thing?

Obviously, ① implies ② in all cases for if it didn't there would be at least one natural number, 𝑦, which may be used to enumerate at least one infinite ordinal, 𝑧.

But does ② imply ①? If ② doesn't imply ① then there must be sets which can't be be placed side-by-side with any infinite ordinal and yet can't be placed side-by-side with any finite ordinal. In short, there must be sets which can't be well-ordered. But the axiom of choice says all sets may be well-ordered, even if it doesn't provide a recipe.

Not only does the axiom of choice say all sets can be well-ordered and thus ② implies ①, but assuming ② implies ① is equivalent to the axiom of choice. It was invented by Zermelo for this purpose and so that ①, ②, and 6 alternative definitions of "finite set" all mean the same thing.

The axiom of choice is basically saying the border between finite and infinite has nothing trapped in it.

#AxiomOfChoice #OrdinalNumbers #SetTheory #FiniteSet

Show threadRichard PennerFeb 16

So we have { 1, 2 } ∈ Fin₁ ⊆ Fin₁ₐ ⊆ Fin₂ ⊆ Fin₃ ⊆ Fin₄ ⊆ Fin₅ ⊆ Fin₆ ⊆ Fin₇

If a set is in Fin₁ then it is considered finite by all the other definitions.

But since the axiom of choice is equivalent to saying every set can be well-ordered, if we accept it VII-finite sets are equinumerous with a finite ordinal and so Fin₇ ⊆ Fin₁ and so the differences between these definitions collapse and ZF becomes ZFC, which is a widely accepted basis for Set Theory.

My consultation with math resources was inspired by a blog post:

https://www.infinitelymore.xyz/p/what-is-the-infinite

#Metamath #ZFC #SetTheory #AxiomOfChoice #FiniteSet #Infinity

What is the infinite?

What does it mean, exactly, to say that a set is infinite? Can we provide precise mathematical definitions of the finite and the infinite?

Infinitely More
Show threadRichard PennerFeb 16

Also from #Metamath I learned #infinity is hard to think about.

A . Lévy in "The independence of various definitions of finiteness" Fundamenta Mathematicae, 46:1-13 (1958) established 8 distinct set-theoretic definitions of a #finiteSet which in ZF cannot be equated without the #AxiomOfChoice

I-finite -- equinumerous with a finite ordinal. // i.e. admits a finite well-order (Numerically Finite) ⟺ the powerset of its powerset is Dedekind finite ⟺ every collection of its subsets has a maximum element ⟺ every collection of subsets has a minimal element

Ia-finite -- not the union of two sets which are not I-finite

II-finite -- every possible way of finding within the set a chain of nested subsets always contains a maximum element (Tarski finite) ⟺ equivalently every such chain contains its intersection ⟺ (Linearly Finite) ⟺ (Stäckel Finite)

|||-finite -- It's powerset is |V-finite finite, (weakly Dedekind finite) ⟺ cannot be mapped onto ordinal ω ⟺ doesn't contain a chain of subsets which can be placed in order with ω

|V-finite -- doesn't have a proper subset which is equinumerous to itself. (Dedekind finite) ⟺ there is no 1-1 map from ordinal ω to it ⟺ it is strictly dominated by the disjoint sum of it and a singleton (acts finite under successor)

V-finite -- it is either empty or strictly dominated by the disjoint sum of it with itself (acts finite under addition)

VI-finite -- it is either empty, a singleton or strictly dominated by the Cartesian product of it with itself (acts finite under multiplication)

VII-finite -- it cannot be infinitely well-ordered (not equiinumerous with the ordinal ω or any larger ordinal)