# A question regarding the Continuum Hypothesis

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

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
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