Free Abelian Groups .... Aluffi Proposition 5.6
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The factor group of the free abelian group generated by all real numbers by the subgroup generated by all relations ##(p,q) = \{p \text{ and } q \text{ are represented by the same Dedekind cut}\}##.lavinia said:So what is the structure of Ras an abelian group under addition?
Edit: Maybe not a satisfactory algebraic explanation but we have to define the relations which makes to real numbers equal.
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I do not see why this is true.fresh_42 said:The factor group of the free abelian group generated by all real numbers by the subgroup generated by all relations ##(p,q) = \{p \text{ and } q \text{ are represented by the same Dedekind cut}\}##.
Edit: Maybe not a satisfactory algebraic explanation but we have to define the relations which makes to real numbers equal.
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Let ##C = ℝ## be the continuum and basis of the free abelian group ##F##. Further let ##ι : C → F## be the embedding (of the alphabet) and ##id: C → ℝ_+## the identity. Then there is a unique group homomorphism ##π: F → ℝ_+## which extends ##id##, i.e. ##id = π \, ι## because ##F## is free. ##π## is onto. The question is what kernel ##π## has or what makes ##4 + 5 = 9## since ##4 + 5## and ##9## are different elements in ##F##. The Dedekind cuts were what occurs to me. Maybe there is a more algebraic way to define the equivalence classes.
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mike is emil's son. (artin)
I will go out on alimb in such fine company and offer my free course notes as well, math 843 on this page:
http://alpha.math.uga.edu/~roy/
Of course Iearned from the sources you have in hand and even cribbed from some of them.
I will go out on alimb in such fine company and offer my free course notes as well, math 843 on this page:
http://alpha.math.uga.edu/~roy/
Of course Iearned from the sources you have in hand and even cribbed from some of them.
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... and a prof of mine once said: A mathematician's talent is transmitted to his son-in-lawmathwonk said:mike is emil's son. (artin)
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It seems to me that Dedekind cuts assume that you already know what addition is.fresh_42 said:Let ##C = ℝ## be the continuum and basis of the free abelian group ##F##. Further let ##ι : C → F## be the embedding (of the alphabet) and ##id: C → ℝ_+## the identity. Then there is a unique group homomorphism ##π: F → ℝ_+## which extends ##id##, i.e. ##id = π \, ι## because ##F## is free. ##π## is onto. The question is what kernel ##π## has or what makes ##4 + 5 = 9## since ##4 + 5## and ##9## are different elements in ##F##. The Dedekind cuts were what occurs to me. Maybe there is a more algebraic way to define the equivalence classes.
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I thought one needs the order in ℝ and the embedding of ℚ. But you are right, this breaks the definition requirements only down on ℚ.lavinia said:It seems to me that Dedekind cuts assume that you already know what addition is.
So we are left with the interesting question: What is addition? Why are 4+5 and 9 equivalent? I have to think about it. I never really lost a thought on it. (PF is an ever lasting fount of challenges ...)
Edit: How about defining ##(a,0) \sim a## for all ##a \in ℝ## and then ##(a,b) \sim c ⇔ \nexists d : ( (a,b) \sim d) ## proceeding by transfinite induction on the well-ordering of ##ℝ##? I'm almost certain that it cant't be done without the axiom and the ordering. But I have to admit that I'm no logician.
Edit2: Perhaps some more care with the definition is needed which I consider a technical issue (well definition, commutativity, inverse).
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As for the addition structure on ##\mathbb{R}##. It might help to look at ##\mathbb{R}## as a ##\mathbb{Q}##-vector space of dimension ##2^{\aleph_0}##. So we could see it as something like ##\bigoplus_{2^{\aleph_0}~\text{factors}} \mathbb{Q}##
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micromass said:As for the addition structure on ##\mathbb{R}##. It might help to look at ##\mathbb{R}## as a ##\mathbb{Q}##-vector space of dimension ##2^{\aleph_0}##. So we could see it as something like ##\bigoplus_{2^{\aleph_0}~\text{factors}} \mathbb{Q}##
Right. Then one needs the structure of the rationals.
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