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[itex]\mathbb R[/itex] can be defined as "any (Dedekind-)complete ordered field". This type of abstract definition is a different kind than e.g. the "equivalence classes of Cauchy sequences" construction. I prefer abstract definitions over explicit constructions, so I would be interested in seeing similar definitions for [itex]\mathbb N,\mathbb Z,\mathbb Q,\mathbb C[/itex]. I think I know the answer for [itex]\mathbb Q[/itex] and [itex]\mathbb C[/itex], so my main concerns are [itex]\mathbb N[/itex] and [itex]\mathbb Z[/itex].
I'm familiar with the following definition of a structure based on Peano's axioms:
Suppose S is a unary operation on a set N, and that 0 is an element of N. The triple (N,S,0) is said to be a Peano system (let me know if that's not the appropriate term), if
(1) S is injective
(2) For all x in N, S(x)≠0
(3) If [itex]0\in E\subset N[/itex] and [itex]S(x)\in E[/itex] for all x in E, then E=N.
Just like with the real numbers, it's possible to show that all structures of this sort are isomorphic, so that it doesn't matter which one of them we call "the natural numbers".
Now how do we include the two binary operations and the total order? Do we have to construct them explicitly from the successor function S, or is there a different way to do it? How about this:
Suppose S is a unary operation on Z, that + and · are binary operations on Z, that 0 is a member of Z and that < is a binary relation on N. The 6-tuple (Z,S,+,·,0,<) is a "foo" (a term I just made up because I don't know if there is a term for this), if
(Edit: I have changed this twice after I posted. First I got some of my ideas for the natural numbers and the integers mixed up, and then I made it more complicated than it needs to be. The structure I'm defining here is supposed to be [itex]\mathbb Z[/itex]).
(1) (Z,+,·) is an abelian ring (with an identity element).
(2) < is a strict total order.
(3) [itex]\big(Z-\{n\in Z|n<0\},S,0\big)[/itex] is a Peano system.
(4) S(0)=1.
Then we prove that all foos are isomorphic (I haven't tried to do this yet) and decide to call the members of a foo (any foo) "integers".
Have you seen a definition like this in the literature?
(And yes, I'm familiar with the construction [itex]0=\emptyset,\ S(n)=\{n\}\cup n[/itex] for the natural numbers. I haven't studied the details yet, but at least I own a good book that covers that very well).
I'm familiar with the following definition of a structure based on Peano's axioms:
Suppose S is a unary operation on a set N, and that 0 is an element of N. The triple (N,S,0) is said to be a Peano system (let me know if that's not the appropriate term), if
(1) S is injective
(2) For all x in N, S(x)≠0
(3) If [itex]0\in E\subset N[/itex] and [itex]S(x)\in E[/itex] for all x in E, then E=N.
Just like with the real numbers, it's possible to show that all structures of this sort are isomorphic, so that it doesn't matter which one of them we call "the natural numbers".
Now how do we include the two binary operations and the total order? Do we have to construct them explicitly from the successor function S, or is there a different way to do it? How about this:
Suppose S is a unary operation on Z, that + and · are binary operations on Z, that 0 is a member of Z and that < is a binary relation on N. The 6-tuple (Z,S,+,·,0,<) is a "foo" (a term I just made up because I don't know if there is a term for this), if
(Edit: I have changed this twice after I posted. First I got some of my ideas for the natural numbers and the integers mixed up, and then I made it more complicated than it needs to be. The structure I'm defining here is supposed to be [itex]\mathbb Z[/itex]).
(1) (Z,+,·) is an abelian ring (with an identity element).
(2) < is a strict total order.
(3) [itex]\big(Z-\{n\in Z|n<0\},S,0\big)[/itex] is a Peano system.
(4) S(0)=1.
Then we prove that all foos are isomorphic (I haven't tried to do this yet) and decide to call the members of a foo (any foo) "integers".
Have you seen a definition like this in the literature?
(And yes, I'm familiar with the construction [itex]0=\emptyset,\ S(n)=\{n\}\cup n[/itex] for the natural numbers. I haven't studied the details yet, but at least I own a good book that covers that very well).
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