Richard PennerTake a set, π₯. Is it finite or infinite? Well what definition do you use?
β βπ¦ β Οβ π₯ β π¦ ; There is a natural number, π¦, such that you may may map π¦ one-to-one onto the set π₯, enumerating each of its members. So π₯ is finite for the same reason { 1, 2, 3 } is finite.
β‘ Β¬ βπ§ β (On β Οβ) π₯ β π§ ; There is no such infinite ordinal, π§, such that you may map π§ one-to-one onto the set π₯. So π₯ is finite because Οβ, the smallest infinite ordinal, cannot be mapped 1-to-1 into it.
Do β and β‘ say the same thing?
Obviously, β implies β‘ in all cases for if it didn't there would be at least one natural number, π¦, which may be used to enumerate at least one infinite ordinal, π§.
But does β‘ imply β ? If β‘ doesn't imply β 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 β‘ implies β , but assuming β‘ implies β is equivalent to the axiom of choice. It was invented by Zermelo for this purpose and so that β , β‘, 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.
Bodling
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