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Given that the order type of the reals $(\mathbb R,<)$ has no definable points, can it be proven that $|\mathbb R|= \aleph_1$ by virtue of the fact that every uncountable proper subset $S$ of $\mathbb R$ (call it $(S,<)$) is (or at least seems to be) indiscernible from $(\mathbb R,<)$ (one could interpret $a as '$a$ is to the left of $b$' or '$b$ is to the right of $a$' and $a=b$ as '$a$ is in the same position as $b$')? Given that $(\mathbb Q,<)$, the order type of the rationals, also has no definable points, what reason (if any) would one want to assume that any completion $(\mathbb R,<)$ of $(\mathbb Q,<)$ by virtue of 'Dedekind cuts' definable in the language of $(\mathbb Q,<)$ would be (in lieu of forcing) other than the minimal completion of $(\mathbb Q,<)$, which would again make $|\mathbb R|= \aleph_1$? Also, how could one 'fatten' $(\mathbb R,<)$ via forcing?

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    Are you trying to *prove* the continuum hypothesis?2012-01-29
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    ...a2012-01-29
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    No, I too hold that CH is independent from ZFC but also understand that CH holds for Closed sets of reals, Open sets of reals and Borel sets of reals. All I am asking is that for (R,<), (Q,<), (R,<) the completion of (Q,<), since (R,<), (Q,<) have no definable points, whether this fact would make CH hold for (R,<). If not, please provide a counterexample (generic or otherwise) in which |R|>aleph-one.2012-01-29
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    I'm not really following your question. $(\mathbb Q,<)$ is an elementary submodel of $(\mathbb R,<)$, the completion is not definable in first order of the language. You can talk about saturation of the model (that is every bounded countable set has a supremum) and then there is a (possibly different) question in your post. Either way, I recall several people working on such problem with Shelah last year, I'll try to find some references.2012-01-29
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    The continuum hypothesis "holds for Borel sets of the reals" means that every Borel set is either countable or the same size as the reals, *not* that their size is $\aleph_1$. This might explain the confusion.2012-01-29
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    Henno: That is what I mean when I say that CH might hold for (R,<). Since (R,<) has no definable elements any uncountable subset S of R under the order '<' would essentially be a copy of R. Since by the Axiom of Choice2012-01-30
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    the notions of infinite and Dedekind infinite are equivalent and because every Dedekind infinite set has a countable subset (in theory one should be able to prove that that countable subset has the order type (Q,<)) every subset of (R,<) will be countable or the same size as R, and CH holds for (R,<). Are you saying that one can have models of set theory i.e. ZFC where |R|=aleph-two, say and every subset of R will either be countable or of size aleph-two? Under what conditions, then will Scott's version of CH implies |R|= aleph-one?2012-01-30

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