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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 $a<b$ 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? – Asaf Karagila Jan 29 '12 at 8:55
...a<b 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'). Since (Q,<) also has no definable points and since (R,<) is the completion of (Q,<) by virtue of 'Dedekind cuts' 'definable in' the language of (Q,<),why would one want to assume that the completion (R,<) of (Q,<) would be other than the minimal completion of (Q,<), again making |R|=aleph-one? How would one go about 'fattening' (R,<) by forcing, since (R,<) has no definable points? – Thomas Benjamin Jan 29 '12 at 8:58
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. – Thomas Benjamin Jan 29 '12 at 9:16
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. – Asaf Karagila Jan 29 '12 at 10:10
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. – Henno Brandsma Jan 29 '12 at 10:34

I believe that your confusion, as Henno points out in his comment, is due to the term "CH holds for Borel sets".

The continuum hypothesis does not say "The continuum has size $\aleph_1$." but rather it says "If $A$ is an uncountable set of real numbers, then $A$ is of size continuum". Note that in the latter there is no specification of the size of the continuum at all. Assuming the continuum can be well ordered then it has the cardinality of some $\aleph$, in which case the statements are equivalents. However it is possible to have the continuum hypothesis to hold without well-ordering the continuum!

The sentence "CH holds for X" means that a member of the family X is either countable or has the cardinality of the continuum. Note that $\mathbb R$ is of course a Borel set, so it say that the Continuum Hypothesis holds for Borel sets effectively says it is true for $\mathbb R$.

Now, the Dedekind completion is unique, that is if there is a linearly ordered set $(A,<)$ then up to isomorphism there is only one completion of it. In the case of $\mathbb Q$ this is indeed $\mathbb R$. Note that such completion is necessarily second order (in the language of linear orders, at least) and therefore not definable internally. In the language of set theory, one can define the completion in first order (simply because we only talk about sets).

This means that every $S\subseteq\mathbb R$ such that $\mathbb Q\subseteq S$ has the exact same Dedekind completion, that is $\mathbb R$. In fact one need not require that $\mathbb Q$ is a subset of $S$ but only that it is dense and unbounded.

Lastly, the question regarding forcing: recall that the cardinality of $\mathbb R$ is exactly the cardinality of $\mathcal P(\mathbb N)$. This is provable from ZFC therefore it will be true in every model, even if the model has many subsets. By forcing we can create a lot of new subsets of $\mathbb N$, in such way that we essentially fatten the real numbers.

How does the idea works? Let us review how we add one subset, the general idea is essentially the same.

The Basic Idea Behind Cohen Forcing or: How I Learned To Stop Worrying And Love Forcing.

We start by a model of set theory, let us call that model $V$. In $V$ we have the natural numbers, often denoted by $\omega$ as the set of finite ordinals. We now define a partially ordered set $(P,\le)$. This set is called a notion of forcing, and this only tells us that it is a partially ordered set and that we say that $p\le q$ means that $p$ is stronger than $q$.

What will be in this sort of partial order? We will approximate the new subset of $\omega$ by functions, and a mysterious being known as a generic filter will then assure to us that this set will exist in a larger universe.

In the case of Cohen forcing we consider $p\in P$ as a function from a finite domain of $\omega$ into $\{0,1\}$. We say that $p\le q$ when $q\subseteq p$, that is $p$ extends $q$ as a function.

We say that $p,q\in P$ are compatible if there is someone stronger than both of them at the same time, in our case it means that $p\cup q$ is a function (so they both agree on their common domain).

Lastly, we say that $D\subseteq P$ is dense if for every $p\in P$ there is $q\in D$ such that $q\le p$, and $G\subseteq P$ is a filter if all its elements are pairwise compatible and if $p\in G$, and $p\le q$ then $q\in G$.

If $G$ is a filter we say that it is generic over $V$ if it intersects with every dense set which lives inside $V$ (this is an important distinction since it is possible that there are dense sets that $V$ does not know about). Let us see what a generic filter does to $P$:

If $G$ is a generic filter, then it intersects every dense set, since we can define for every $n\in\omega$ a dense set: $D_n=\{p\in P\mid n\in Dom(p)\}$, why is it dense? If $p\in P$ either $n$ is in its domain and it is in $D_n$ or it isn't then $p\cup\{\langle n,1\rangle\}$ is a stronger condition which lies in $D_n$.

That means that for every $n\in\omega$ we can find $p\in G$ such that $n\in Dom(p)$. Furthermore, $G$ is a filter so every two elements in $G$ are compatible. This means that $\bigcup G$ is a function, otherwise we could find two elements of $G$ which are not compatible.

This means that from $G$ we can define a subset of $\omega$. If $G$ was not in $V$ then this subset is not in $V$, because given the subset we can easily reconstruct this filter. On the other hand it is not possible that the subset we defined was already in the universe since given a filter $F\in V$ the set $P\setminus F$ is dense, therefore the generic filter is new to the universe.

We have, if so, a brand new set added to $\omega$, which means that the new universe $V[G]$ has another real number that was not in the original one.

The mysterious generic filter: Of course the fact that such generic filter exists is impossible to prove. However if $V$ is countable we can enumerate the dense sets (there are only countably many of those) and then define from this enumeration the generic filter. This is not the only way to claim it can exist, there are additional axioms and there is the Boolean-valued approach to forcing (which is way too long to write in this already-too-long answer) from which we can deduce that such filter exist.

Now to show that it is possible to have $|\mathbb R|=\aleph_2$ we simply add $\aleph_2$ many real numbers, we can consider those as functions from $\omega_2\times\omega$ into $\{0,1\}$ whose domain is finite. Of course that we have to argue something else in such case: We did not add a bijection from $\omega_1$ to $\omega_2$. If we would have added such bijection then in the new universe the continuum would be larger but will retain its cardinality (the new universe will not know that we have $\aleph_2$ many reals because now $\aleph_2$ would be different).

This is a slightly more technical proof, but we've checked this thoroughly and you can rest assured that this does not happen.

Further Reading Material:

  1. Mathematical statement with simplest independence proof (another relatively basic forcing example)
  2. Generic filter over $V$ (why should we believe that generic filters exist?)
  3. Showing $C=\bigg\{ \sum_{n=1}^\infty a_n 3^{-n}: a_n=0,2 \bigg\}$ is uncountable (bijection between $\mathcal P(\mathbb N)$ and $\mathbb R$ follows from this + Dedekind cuts + Cantor-Bernstein theorem)
  4. Relationship between Continuum Hypothesis and Special Aleph Hypothesis under ZF (The continuum hypothesis can hold without $2^{\aleph_0}=\aleph_1$).
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So essentially you are saying that for (R,<) (which has no definable elements) forcing would add extra points to the point set R? Also, as regards R being a Borel set, since CH was proven for Borel sets by Alexandrov in 1916, why was this proof not regarded as a proof of CH for R if in fact R is a Borel set? Indeed,since one can put R in a 1-1 correspondence with the elements of the open interval (0,1) and CH holds for (0,1), why doesn't that constitute a proof of CH for R? – Thomas Benjamin Jan 30 '12 at 6:56
I guess what concerns me is that though forcing is good at constructing 'toy' (countable) models of ZFC,what do the toy models say about what properties the 'real universe' V must have? – Thomas Benjamin Jan 30 '12 at 7:04
@Thomas, read the second line in my answer. When we say that CH holds for Borel sets it means that there is no Borel set which is of cardinality strictly between $\aleph_0$ and $2^{\aleph_0}$, regardless the possible fact of the latter being $\aleph_1$ or not. – Asaf Karagila Jan 30 '12 at 7:44
There is a very delicate point about countable models. Sometimes they are used to prove things about the universe (in many different ways) and sometimes you just assume that you're living in one - since if ZFC is consistent then it has a model, and therefore a countable model. – Asaf Karagila Jan 30 '12 at 7:46
Also, when you say of (Q,<), (R,<) is the unique completion, don't you mean to say (R,<) is the completion not obtained by forcing. If that is the the case then can't one assume the completion (R,<) of (Q,<) is just the minimal one, in which case |R|=aleph-one? – Thomas Benjamin Jan 30 '12 at 8:43

Here is something that you may find interesting but seems to go against your proposed approach to CH.

First, PFA, the Proper Forcing Axiom, is a very useful and a very strong hypothesis independent of the usual axioms. It proves that the continuum has size $\aleph_2$ and that

every two dense subsets of ${\mathbb R}$ of size $\aleph_1$ are order isomorphic.

This is a result of the late J. Baumgartner, and can be found in his nice paper Applications of the proper forcing axiom, in the Handbook of set-theoretic topology, K. Kunen, J. Vaughan eds., 913–959, North-Holland, Amsterdam, 1984.

On the other hand, it is a theorem of Sierpiński that the displayed statement fails spectacularly in the presence of CH: In W. Sierpiński, Sur un problème concernant les types de dimensions, Fund. Math. 19 (1932), 65–71 (available here), the following is proved:

If CH holds, then there are $2^{\aleph_1}$ non-isomorphic dense subsets of reals of size $\aleph_1$.

In fact, what Sierpiński proves is that any subset of ${\mathbb R}$ of the same size as the reals contains a subset $Y$ of the same size as the reals and such that any monotone $f:Y\to Y$ differs from the identity in a set of size less than $|{\mathbb R}|$.

It seems that this indicates that an indiscernibility argument along your lines is in fact an argument against CH (as CH proves there are many non-order isomorphic subsets of ${\mathbb R}$ of size $\aleph_1$, and I do not see how to formalize your indiscernibility argument in a way that does not entail order-isomorphism).

It also seems that such an argument would indicate that a strong negation of CH (something that gives a result like the consequence of PFA displayed above) would be more desirable, since it recovers part of your suggestion (all dense subsets of an appropriate size are isomorphic). Of course, this is a weakening of your approach, as it is impossible to have this if we insist that the sets must have the same size as ${\mathbb R}$.

It is actually clear nowadays that in many situations, and this extends well beyond the theory of orders and includes examples from functional analysis and other fields, PFA provides us with nice classification results, while CH leaves us with non-classifications, where the number of non-isomorphic objects of the desired kind is as large as possible.

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Baumgartner's result is just some work added to Shelah 100, no? I recall about a year ago that some distinguished guests came to Israel and my advisor met with them to talk about this result and I recall some relation to PFA. In the original thread this is really the result I wanted to cite before Henno gave his comment that made me change my mind and write my answer. – Asaf Karagila Jan 29 '12 at 21:42
It goes well beyond Shelah 100, I would think. It is one of the first results showing that PFA was worth investigating beyond a technical strengthening of MA. – Andrés E. Caicedo Jan 29 '12 at 21:56
Well, Shelah 100 did come out a couple of years prior to PFA's formulation I believe (published in 1980). – Asaf Karagila Jan 29 '12 at 21:59
There are some interesting stories around the 100 paper and getting it published, but probably you have heard them all by now. – Andrés E. Caicedo Jan 29 '12 at 22:04
I have heard none of them! I will be in Jerusalem tomorrow so I'll snoop around and ask for the folklore. – Asaf Karagila Jan 29 '12 at 22:06

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