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I have not dealt professionally with set theory, so excuse me if my way of formulating this question does not completely follow standard terminology. Actually, my question is about whether or not the following idea can be formalised:

The question is whether it makes sense to talk about a model for ZFC where the number of real numbers is maximal. I know that Paul Cohen has proved that there exist models for ZFC where $\Bbb R$ has an arbitrarily large cardinality. However, as Löwenheim–Skolem shows us, cardinality within a model is not the same as the cardinality of the number of elements in a model. So would it be possible to construct a model $M$ for ZFC in which, for any model $M'$ and a real number $r'$ in $M'$, there exists a real number $r$ in $M$ which has the same numerical value, i.e. the same digits?

The reason I would think this could actually be the case is this: We choose to construct the real numbers $\Bbb R$ as the set of maps $r\colon\Bbb Z\to\{0,1,2,\ldots, 9\}$ with $r(n) = 0$ for all sufficiently large $n$. Intuitively, $r_n:=r(n)$ is the $n$th digit of $r$, i.e. $r = r_N\ldots r_0.r_{-1} r_{-2}\ldots$ for some $N\ge 0$. Of course we shall have to apply an equivalence relation because, for instance, $0.999999\ldots$ must be equal to $1$. Now, this is where it goes vague, but what we do is to include any such sequence of numbers in $\{0,1,2,\ldots,9\}$. I mean literally that any such combination must correpsond to a well-defined map $\Bbb Z\to \{0,1,2,\ldots, 9\}$.

But is this idea of "any combination" possible to formalise in model theory? In this case, is my maximal model for $\Bbb R$ possible to construct formally?

Gaussler
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    It's a troublesome question. If you know enough set theory, you can probably answer it yourself; if you don't then many of the delicate points are often lost on you and any proper answer will be lost on you. I hope that I managed to make some sense in my answer, but I fear that I have not. – Asaf Karagila Jul 28 '14 at 20:46

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What you seem to be really asking is whether or not there is a model which includes all the real numbers.

Under certain, perhaps even reasonable, assumptions yes. There is such model. Note that this model, however, must be uncountable. And since we want set-models, class models (the universe of sets, for example) are out of the picture. Otherwise, $V$ itself is such model.

As for "construct", since we cannot construct any models of set theory, the term is a bit vague. When we construct a model of $\sf ZFC+\lnot CH$ we begin with a model of $\sf ZFC$, we don't just make one out of thin air. But here you require more, and so the starting conditions require more.

If there exists an inaccessible cardinal $\kappa$, for example, then $V_\kappa$ is a model of $\sf ZFC$ which includes all the real numbers.

Finally, let me point out that in different models of $\sf ZFC$ real numbers can behave differently. For example, if there is any model of $\sf ZFC$ then there is such $M$ such that $\{x\mid M\models x\text{ is an integer}\}$ is an uncountable set, as large as you'd like it to be. If $M$ has more integers than the universe has real numbers, then there are real numbers in $M$ -- or rather objects which $M$ thinks of as real numbers, which cannot be matched by actual real numbers.

So instead the question should be reduced to models which agree with the universe about what are real numbers. Or to make it simpler, to transitive models of $\sf ZFC$. And in this case, there are either such models or there are not. Depending on your starting assumptions.

Asaf Karagila
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    I see I definitely shouldn't have used the word "construct," because as you point out, construction is out of the question, and it actually isn't what I was focusing on. What I meant by "construct" was more like "formally define." What I was thinking about was something like: Take any model $M$; if there exists another model $M'$ with a real number $r'$ which does not correspond to a real number i $M$, have it added as $r$ in $M$. Continue for all models. But then I end up taking the union of all real numbers in all models, which sounds like something which could be highly controversial. – Gaussler Jul 29 '14 at 07:04
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    But how would you "add it to the model"? And when you do, why would it be a real number in the new model? – Asaf Karagila Jul 29 '14 at 07:10
  • Can't you do something like adding the map I mentioned earlier and then add all sets that ZFC guarantees must exist as a consequence of the existence of this map. That should be a new model for ZFC. (I do realise that this may not make much sense, but as mentioned, I'm new to set theory.) – Gaussler Jul 29 '14 at 07:17
  • Concretely, given a map $r'\colon\Bbb Z\to{0,1,\ldots,9}$ in $M'$, add a map $r$ in $M$ taking the corresponding values. By the construction above, this should be a new real number. – Gaussler Jul 29 '14 at 07:19
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    Do you see why this is not possible when $M$ and $M'$ have different sets as their integers? – Asaf Karagila Jul 29 '14 at 07:26
  • I'm sorry to admit it, but I don't really see this problem. For there is an easy one-to-one correspondence between $0,1,\ldots,9\in M$ and $0',1',\ldots,9'\in M'$. Of course, this is a meta-model correspondence (if there is such a thing). But yes, I see that I may be getting out in deep water on this one. ;-) – Gaussler Jul 29 '14 at 07:43
  • This is why I said that $r$ must take "the corresponding values," not "the same values." – Gaussler Jul 29 '14 at 07:44
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    But there are no "corresponding values". How can there be if one model has more integers than the second model has sets? (For example...) I think that the issue is that you don't understand why having different sets of integers means that the real numbers are different. Think about it. Long and hard. – Asaf Karagila Jul 29 '14 at 07:48
  • I shall try, thank you for your help. :-) – Gaussler Jul 29 '14 at 07:59
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    Yeah, okay, I realise you're talking about $\Bbb Z$ being different. I thought you meant ${0,1,2,\ldots, 9}$. Don't really see why I didn't catch it. Thank you for the explanation. :-) – Gaussler Jul 29 '14 at 10:52