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Why can't the well ordering of the reals be proven with Dedikind cuts without AC? 
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#19
Apr610, 11:51 AM

P: 424

See the wellordering theorem on Wikipedia. Basically you form the set of all wellordered subsets of the real numbers, then you invoke Zorn's lemma to conclude that there is a maximal such wellordered subset, and you then note that it must actually be a wellordering of the real numbers themselves.



#20
Apr610, 02:34 PM

P: 41

Anyway using AC one can proves that every set can be wellordered and therefore has cardinality equal to some [tex]\aleph_{\alpha}[/tex] To SW VandeCarr, AC equivalent to wellordering theorem, and from it every set can be wellordered, hence set of cardinality [tex]2^{\aleph_0}[/tex] can be well ordered. So with AC we can compare cardinalities. Definition of AC: Every family of nonempty sets has a choice function. Choice function: If S is a family of sets and [tex]\emptyset \not \in S[/tex] then a choice function for S is f on S such that [tex]f(X) \in X[/tex] for every [tex]X \in S[/tex]. THM: AC > every set can be wellordered. Proof: Let A be a set. For every [tex] \alpha[/tex] we can define [tex] a_{\alpha} = f(A  \{a_{\xi}: \xi < \alpha\}) [/tex] if [tex] A  \{a_{\xi}: \xi < \alpha\}[/tex] is not empty, by induction. let [tex] \theta[/tex] be the least ordinal such that [tex] A = \{a_{\xi} : \xi < \theta\}[/tex]. So we found enumeration of A. THM: Everyset can be wellordered > AC Proof: Let S be a family of sets then we shall wellorder [tex] \cup S[/tex] and f(X) will be defined as the least element of X for every [tex]X \in S[/tex]. Now since reals is [tex] 2^{\aleph_0} [/tex] then it is can be wellordered and hence be one of alephs. 


#21
Apr610, 03:36 PM

P: 2,501




#22
Apr610, 04:19 PM

P: 41

Anyway, idea of the proof is simple. Choose one element after another and by doing so you order them. 


#23
Apr610, 05:05 PM

P: 2,501

While I won't pursue the issue here, it seems we could proceed with some kind of constructive approach for the reals with the same logic: constructing the next element ad infinitum by continuation. I don't have a problem using AC in conjunction with other nonequivalent axioms/theorems, but I don't like proofs that essentially rely on AC alone. Thanks for clarifying this for me. 


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