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Background: We know that PA has more models than the intended model, N, because it is not strong enough and is also satisfied by non-intended models, known as non-standard models of arithmetic. When we talk about the standard model N, I somehow assume that there is some way to characterize it without ambiguity, so it is well defined and can be identified as the only model that is isomorphic to the natural numbers that we use everyday. Every statement of PA has a specific truth value on N (either true or false, regardless of our ability to know the answer). But I am not sure if that is also the case for the real numbers. Informally, they are a value that represents a quantity along a continuous line. Also, they can be defined axiomatically up to an isomorphism in different ways. They are also been shown to "fill" the real line, so there are no more numbers than them on it.

Question(s): Each statement about the naturals has a specific truth value: Is this the same case for the reals? are they defined with such precision? My doubt comes from the fact that there are models of set theory in which the CH is true and others in which it is false. Do this mean that the actual reals do not have a specific truth value for the CH, or does it mean that they have a specific value (which we don't know yet), and that models with a different value are models of non-standard reals?

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I have seen people use [model-theory] when they have meant to use [mathematical-modeling] before, but never the other way around! – Asaf Karagila Feb 13 '13 at 21:49
sorry, I don't know what tags to use, please feel free to edit them – Wolphram jonny Feb 13 '13 at 21:50
@julianfernandez Someone will probably give a proper answer soon, so I'll give you a half-ass answer in the meantime: the CH is independent from the ZFC axioms, meaning you can assume it is true, or assume it false and still have a consistent theory, that is, no contradictions can be derived from the axioms by assuming either CH or $\neg$ CH, within ZFC. – Git Gud Feb 13 '13 at 21:55
@GitGud, thanks Git, I am aware of that, and the same can happen in PA (I mean, have two consistent theories with opposite values of an independent statement from PA, but only those theories in which the truth value agrees with the one of the naturals represent a model of the standard natural numbers – Wolphram jonny Feb 13 '13 at 22:03
If $\mathsf{CH}$ fails, we can go to a forcing extension where it holds, and no reals have been added. I would take this as evidence that $\mathsf{CH}$ is not quite an intrinsic question about the natural model of the reals. – Andrés E. Caicedo Feb 13 '13 at 22:36
up vote 8 down vote accepted

The real numbers, much like the natural numbers, have a canonical second-order model. This means that given a model of set theory there is just one model up to isomorphism.

Of course this model can change if we change the model of set theory, but so can the standard model of the natural numbers. That is if $M$ and $N$ are different models of ZFC they might have different collections of what they perceive as natural numbers or as real numbers.

When we talk about the real numbers as a first-order theory we often consider the theory of real closed fields. We can consider models of this theory which are non-standard. For example the hyperreal field is an example of such model, and non-standard models are useful for non-standard analysis. What makes them non-standard? Well, they usually have "infinite" numbers, which are numbers that are larger than any finite repetition of adding $1$ to itself, much like how non-standard integers exist in non-standard models of PA.

However CH has nothing to do with this. Because CH is not something we can really formulate in the language of ordered fields. In this theory we can't really say one set has a larger cardinality than another. This would be equivalent to asking whether or not there are non-standard models of the real numbers because there are non-abelian groups, and so being an abelian group is independent of group theory (which the real numbers are a model of, of course).

It is true that within a fixed model of ZFC the standard model of the natural numbers is always countable; but it is also true that the real numbers always have cardinality equal to the power set of the standard model of the natural numbers, that is, $|\mathbb R|=2^{\aleph_0}$, regardless to it being $\aleph_1$ or $\aleph_{50043}$. In fact this is true even if the axiom of choice fails and $2^{\aleph_0}$ is not an ordinal at all.

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at Asaf: I am not sure to understand what you mean when you say "if M and N are different models of ZFC they might have different collections of what they perceive as natural numbers or as real numbers". do you mean non-standard models, or do you mean that we do not enough about the standard N, so we dont really know if the intended model N of PA are the "actual" N? I mean, you agree that there is only one model of N, right? and that is either the recursive model of PA (which I believe is independent of any definition set theory) or the categorical model of second order arithmetic (the same?) – Wolphram jonny Feb 19 '13 at 0:06
@julian: Every two standard models of ZFC have the same natural numbers. But non-standard models of ZFC can have non-standard integers. But internally speaking every model sees its integers as "the" standard model. But externally we can have a model whose integers are externally uncountable. As for the real numbers, even in the case of standard models of ZFC the collection of reals can differ (in internal cardinality as well). – Asaf Karagila Feb 19 '13 at 0:09
I think some of the concepts are beyond my reach. In lay terms, would it be correct correct to say that the standard naturals and their properties are unique (each statement has a true or false value), but some questions that involve additional definitions (such as sets) are not really properties of the naturals (they may make sense or not if the definitions are precise enough). So, would it be the same with the real numbers? they are a specific structure, but some questions about them are not precisely well defined (such as the CH which is rather dependent of what do we mean by a set?) – Wolphram jonny Feb 19 '13 at 0:19
@julian: Not exactly. Because often (in set theory) we mean set just as an object in a universe of ZFC (or some other theory). The naturals have their own place in mathematics. The real numbers, however, carry an ordered field structure which is very unique in second-order; and "unique enough" in first-order. But this structure is not really topological, analytic, whatever. It's just the field and the order. Whereas the natural numbers are often coupled with the induction axiom[s] which gives them the expressive strength. [cont...] – Asaf Karagila Feb 19 '13 at 0:34
[...] So while the natural numbers are rewarded with a very strong first-order expressibility, the real numbers have a very weak expressibility powers as an ordered field. If you were to give them a different structure things might be different (e.g. you can add predicate for the natural numbers and the axioms of PA relativized to this predicate). So whereas the natural numbers in PA can express and ask whether or not their theory is consistent, the real numbers are oblivious as an ordered field. [cont...] – Asaf Karagila Feb 19 '13 at 0:37

First you need to realize that given a particular model $M$ of a theory $T$, for any statement in the language holds that the statement is either true in $M$ or false in $M$. There are no other possibilities. In that sense, any fixed model of PA or of the real numbers (under whatever theory you want to designate as the theory of the reals) is unambiguous and it decides each and every statement as either true or false.

Now, things become less clear cut when considering the consequences of the theory $T$ rather than studying a particular model of it. As for PA, as a first order theory it indeed has many nonstandard models. But, if you add to PA the second order statement of induction then it becomes categorical (meaning that every two models of it are isomorphic). A similar phenomenon is true of the real numbers. The first order theory of the reals allows nonstandard models. But, if you add the completeness axiom then it becomes categorical.

Since most people working with the natural numbers accept the principle of induction (much stronger actually, they accept the axiom of choice) there is essentially just one model of the natural numbers. Since most people working with the reals certainly accept the axiom of completeness (and most also accept the axiom of choice) there is essentially just one model of the reals.

It should be noted though the sometimes people deliberately consider nonstandard models, like in nonstandard analysis, for the purposes of the study of very ordinary analysis. It has advantages and disadvantages.

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It's important to point out that the completeness axiom is a second-order axiom. – Asaf Karagila Feb 13 '13 at 22:17
+1 to Asaf Karagila's comment, but maybe this will seem to be qualifying it a bit: If you have a structure including the reals with the usual operations and order, and subsets of the reals, and the membership relation, then you can write a first-order least-upper-bound axiom. But of course it only says that all sets that are members of the model obey the axiom. If you add a new axiom saying all sets are members of the model, then that is a second-order axiom. – Michael Hardy Feb 13 '13 at 23:25
at Ittay: so, if I followed you well, do you mean that the CH has a specific truth value for the categorical definition of the real numbers? – Wolphram jonny Feb 19 '13 at 0:24
no, since CH is independent of the axioms of the real number system. And actually, CH has some consequences for analytical results so it makes a bit of a difference whether you believe in CH or its negation. The categoricity of the second order theory of the reals is within a given model of sets. – Ittay Weiss Feb 19 '13 at 0:27
so, we do not yet know enough about the reals to give them a categorical definition that is independent of our understanding of sets? but that is not the case for N right? you can define N as the only computable (tennembaum's theorem) model of PA, and there is no ambiguity there – Wolphram jonny Feb 19 '13 at 0:35

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