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Consider the following theorem of Cantor:

If one has a simply-ordered set (M.<) which fulfills the three conditions: a) |M| = aleph-null; b) M has no greatest or least element; c) M is everywhere dense; then M has the order-type of the rationals (quoted from Dauben's book GEORG CANTOR,pg 187).

Let (M,<) be a simply-ordered set of points satisfying a,b,and c on which the automorphism

group defined on M acts transitively by translation so that all elements m of M 'look alike'.

I want to consider a completion of (M,<) by 'Dedekind Cuts' C of M to a simply-ordered set

(R,<) which has the following properties:

a') (R,<) is perfect; b') (R,<) contains the set (M.<) previously defined which is so related to (R,<) that between any two elements r(0), r(1) of R, elements of M occur (by another theorem of Cantor, (Dauben, ibid, pg 191) (R,<) has the order-type of the reals--sorry for the possible redundancy here, I hope it is not too annoying).

Given that (M,<) is defined as a set of points on which the automorphism group defined

on M acts transitively by translation (so that all elements m of M 'look alike'), is

the completion (R,<) of (M,<) satisfying a',b' the unique completion and does |R|=

aleph-one? (Again, I am sorry if the question has possible hidden vagueness--people

who might try to answer this question can suggest how to remove these.)

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What does it mean for $\langle R,\lt \rangle$ to be perfect? $\;$ –  Ricky Demer Jan 15 '12 at 11:07
I don’t understand what you want. Given that (M, <) is isomorphic to the rationals, your condition that all elements of M look alike is automatically satisfied, and you can take for (R, <) the ordered set of real numbers (by composing the injection of Q in R by an isomorphism between M and Q). And of course, the cardinal of R has no reason to be $\aleph_1$ unless you accept the continuum hypothesis. –  Guillaume Brunerie Jan 15 '12 at 11:42
@Ricky Demer: I would guess that by pefect he means that the order has no isolated points. @Thomas Benjamin: It is not at all clear what you mean. The completion of the rationals (or in your case, of a countable dense linear order without endpoints) is unique, has size $2^{\aleph_0}$, is isomorphic to the reals, and has the property that every finite partial isomorphism extends to the whole order. –  Stefan Geschke Jan 15 '12 at 13:22
To Guillaume Brunerie: I took the liberty of looking at some of your questions and found your question on forcing interesting. I now have a question for you. Consider the real line as defined in L. Is that 'real line' complete (complete in the sense that in L, every nonempty set of real numbers definable in L that is bounded from above has a least upper bound)? If that is the case, then given a forcing extension of L which adds a Cohen real, what is actually being added to the real line in L when by the forcing argument (construction?) the Cohen real is 'added' to the real line of L? –  Thomas Benjamin Jan 18 '12 at 16:08
@Thomas Benjamin: The real line of $L$ is complete in the sense of $L$, but it is not complete in the sense of the universe that has a Cohen real over $L$. The Dedekind cut defined by the Cohen real is not filled in $L$. But the point is that $L$ also does not contain this cut and this is why it doesn't know about the incompleteness of its real line. –  Stefan Geschke Jan 21 '12 at 11:18

1 Answer 1

up vote -5 down vote accepted

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'm sorry that this website did not print my answer either completely of correctly. the missing parts are {m is a member of M|r(1) < m<r(2)} where r(1), r(2) are members of M is countable, I am using AC in the form "Every set can be well-ordered" and the following alleged equivalence of CH "CH holds iff (R,<) is the union of an increasing chain of countable sets." My apologies for the website. If you have difficulties understanding the truncated answer email me at thantifaxathbeta@gmail.com and I will clarify any difficulties. –  Thomas Benjamin Jan 18 '12 at 11:03
I fixed the display issue. In general, when things like that happen it is because you are using the less-than sign < which can be confused for HTML markup. A judicious replacement of < by the character entity &lt; when typing up your response should be enough to clear the issue. –  Willie Wong Jan 18 '12 at 11:20
Thanks Willie! I sure appreciate that. –  Thomas Benjamin Jan 18 '12 at 15:42
The fact you quote from my paper with Reitz and Linetsky is the (completely trivial) observation that one can move any point to any other by translation in the rational order $\langle\mathbb{Q},\lt\rangle$ and also in $\langle\mathbb{R},\lt\rangle$. In this restricted language, therefore, it follows that there are no definable elements. As one adds structure, however, there are of course many definable elements in the corresponding expanded languages, as well as definable cuts in $\mathbb{Q}$. –  Joel David Hamkins Jan 18 '12 at 15:56
This answer makes no sense to me. One problem is that the notion of "definable" presupposes a language and no language has been specified. Another problem (which might cancel the first) is that definability seems to be just a red herring as far as the question is concerned. Yet the answer seems to make much of the lack of definability. Then there are assertions like "there are no gaps except in relation to its completion" where "in relation to" has no discernible meaning. There are more problems, but I'm not going to start a new comment when I run out of space. Downvoting. –  Andreas Blass Jan 18 '12 at 16:07

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