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I am going to produce a tentative answer to my question to see if someone can see an error with my reasoning. Since (M,<) is isomorphic to the rationals, it would seem that one could define a set of 'Dedekind Cuts' (A(1),A(2)) on (M,<). Following Dedekind, there would be two types of cuts, those produced by elements m of M, and those produced by the 'irrationals'. The difficulty here is that if you believe Hamkins, Linetsky, and Reitz in their paper "Pointwise Definable Models of Set Theory", in (M,<), just as in its completion (R,<), there are no definable elements, since (quoting the paper) "the automorphism group acts transitively by translation and all points look alike". Dedekind, in his essay "Continuity and Irrational Numbers", uses what Hamkins, et al would consider a definable element in the set of cuts as an example of a cut not produced by a rational. Since in (M,<) and in its completion (R,<) there are (by assumption) no definable elements I suppose one would have to identify the 'irrational numbers' 'defined' by (M,<) (hard to do since each of the members m of M are not defined) by the cuts (G(1),G(2)) produced by the 'gaps' in M (of course in (M,<) there are no 'gaps' except in relation to its completion (R,<) but since (R,<) is to be constructed from (M,<) this seems to be unacceptable circular reasoning). Slogging through this allegedly "unacceptable circular reasoning" it is clear that both the cuts produced by the elements m of M and the cuts produced by the 'gaps' 'in M', being partitions of M, a countable set, are themselves countable. Also, by b' (here I will denote a 'gap' by the symbol gap(i) where i is an ordinal) there will be between any two gaps gap(1), gap(2) elements of M, the sets of the type {m is a member of M| gap(1) < m < gap(2)} are countable, just as for the case {m is a member of M| r(1) < m < r(2)} if r(1), r(2) are themselves members of M. Now assuming AC in the form "Every set can be well-ordered and the following alleged equivalence to the Continuum Hypothesis

"CH holds iff (R,<) is the union of an increasing chain of countable sets."

one can seemingly 'prove' |R| of(R,<) is aleph-one by using AC to well-order the elements of M in R and the 'gaps' gap(i) in R and use this well-ordering to form an increasing chain of countable sets, the union of which is R of (R,<) (since both the cuts produced by elements of M and the gaps gap(i) are countable sets and both the types of sets defined by b' where r(1), r(2) are either members of M or gaps gap(1),gap(2) are also countable this argument seems to be correct if the alleged equivalence to CH I used is in fact correct). Since I did not use any forcing argument (at least not to my knowledge) (R,<) is the unique completion of (M,<). Where have I gone wrong (if in fact I have)?

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I am going to produce a tentative answer to my question to see if someone can see an error with my reasoning. Since (M,<) is isomorphic to the rationals, it would seem that one could define a set of 'Dedekind Cuts' (A(1),A(2)) on (M,<). Following Dedekind, there would be two types of cuts, those produced by elements m of M, and those produced by the 'irrationals'. The difficulty here is that if you believe Hamkins, Linetsky, and Reitz in their paper "Pointwise Definable Models of Set Theory", in (M,<), just as in its completion (R,<), there are no definable elements, since (quoting the paper) "the automorphism group acts transitively by translation and all points look alike". Dedekind, in his essay "Continuity and Irrational Numbers", uses what Hamkins, et al would consider a definable element in the set of cuts as an example of a cut not produced by a rational. Since in (M,<) and in its completion (R,<) there are (by assumption) no definable elements I suppose one would have to identify the 'irrational numbers' 'defined' by (M,<) (hard to do since each of the members m of M are not defined) by the cuts (G(1),G(2)) produced by the 'gaps' in M (of course in (M,<) there are no 'gaps' except in relation to its completion (R,<) but since (R,<) is to be constructed from (M,<) this seems to be unacceptable circular reasoning). Slogging through this allegedly "unacceptable circular reasoning" it is clear that both the cuts produced by the elements m of M and the cuts produced by the 'gaps' 'in M', being partitions of M, a countable set, are themselves countable. Also, by b' (here I will denote a 'gap' by the symbol gap(i) where i is an ordinal) there will be between any two gaps gap(1), gap(2) elements of M, the sets of the type {m is a member of M| gap(1) < m < gap(2)} are countable, just as for the case {m is a member of M| r(1) < m

"CH holds iff (R,<) is the union of an increasing chain of countable sets."

one can seemingly 'prove' |R| of(R,<) is aleph-one by using AC to well-order the elements of M in R and the 'gaps' gap(i) in R and use this well-ordering to form an increasing chain of countable sets, the union of which is R of (R,<) (since both the cuts produced by elements of M and the gaps gap(i) are countable sets and both the types of sets defined by b' where r(1), r(2) are either members of M or gaps gap(1),gap(2) are also countable this argument seems to be correct if the alleged equivalence to CH I used is in fact correct). Since I did not use any forcing argument (at least not to my knowledge) (R,<) is the unique completion of (M,<). Where have I gone wrong (if in fact I have)?