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For the sake of brevity, throughout this post I will identify real numbers with subsets of $\mathbb{N}$. The question that I want to ask here is more heuristic than definite; I want to understand the deeper intuition behind the fact that the Continuum Hypothesis is independent of the ZFC axioms. I am familiar with Godel's and Cohen's proof showing the independence of CH, and now I am trying to figure out the deeper heuristic behind them.

If we compare another famous independence problem, the independence of Euclid's parallel postulate we notice (with great hindsight) that the reason why this postulate is independent is because the concept of 'line' has not been sufficiently well defined. I.e the concept of a line depends on the ambient space you are working in; on a sphere, a line has very different properties than on a plane.

When we transfer this intuition to the Theory of Sets, we see that there is a set theoretic concept which similarly lacks a clear definition, namely that of 'subset'. Take $P(\mathbb{N}) = \mathbb{R}$ for example. All the subsets we have ever encountered are 'definable' in the sense that they can be recursively enumerated by a finite algorithm; e.g.$$\{2,4,6,8,...\}$$ $$\{2,3,5,7,...\}$$ $$\{1,4,9,16,...\}$$ etc... and since an algorithm is just a finite string of symbols, there are countably many 'definable' reals. But we know that $\mathbb{R}$ is uncountable. Therefore there must be real numbers hidden 'deep' within $\mathbb{R}$, which we can call the deep structure of $\mathbb{R}$. Now if my intuition is correct, it is precisely the fact that these 'deep' reals cannot be defined in ZFC that the exact number of them (i.e. $|\mathbb{R}|$) is undecided by the rest of ZFC. I would like to make this notion more definite.

So I ask: Is this the right way to understand the deeper reason behind the independence of CH? Are there any expository articles been written on this subject that I can access? (I am not really interested in things like PFA $\implies 2^{{\aleph _{0}}}=\aleph _{2}$ right now). Can anyone here explain intuitively why the undefinability of most real numbers allows for us to add arbitrarily large numbers of reals to $\mathbb{R}$ (I.e. in Cohen real forcing). Many thanks.

Asaf Karagila
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Elie Bergman
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    Any attempt to compare CH with the Parallels Postulate is essentially lying to yourself. Set theory and geometry are two very different things. This is like trying to compare CH with "Given a field, $F$, is $\sqrt2\in F$?" (or does $F\models\exists x(x\cdot x=1+1)$ in a strictly formal way). This is also independent, but is it because of some definability issues or "ambiguity in defining" of anything? – Asaf Karagila Apr 10 '18 at 13:24
  • I don't think so. The Cantor diagonal series is exactly defined. – Raphael J.F. Berger Apr 10 '18 at 13:25
  • I have no idea what you mean by "hidden deep within R". No real number is "hidden" any more than any other. If, for example, you think of the integers as not "hidden" at all that is entirely an artifact of the particular way we have defined the integers. – user247327 Apr 10 '18 at 13:25
  • To build on user247327's comment, what I believe you are talking about is the uncomputable reals. When you say, "it is precisely the fact that these 'deep' reals cannot be defined in ZFC that the exact number of them is undecided by the rest of ZFC", this is epistemically true. By your own admission, if they were all computable, then they would all be definable in countably many strings of the language of ZFC and so would be, themselves, countable. So... – Isky Mathews Apr 10 '18 at 14:38
  • @AsafKaragila Okay, but we have no idea what a random field looks like, whereas the concept of the Real Line is defined independently of the model of ZFC we are using. Even a 6 year old has a direct intuition of what the Real Line is, therefore it is strange that something that is a well defined definite object should have such enormous indeterminiteness as it varies from one model of ZFC to another. It is as if the axioms of ZFC are complex enough to define R, but not complex enough to define R FULLY (i.e. determine its cardinality). – Elie Bergman Apr 10 '18 at 15:51
  • Elie, fine. Just talk about finite fields. Just talk about $\Bbb Q$ and $\Bbb Q[\sqrt 2]$. Two very concrete fields. Geometry is very simple, as far as theories go. – Asaf Karagila Apr 10 '18 at 15:53
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    Not to mention that CH is also consistent with a lot of other extensions of ZFC which practically tell you that your suggestion of intuition is completely off. There can be only countably many constructible reals, but CH holds. The reals can be pointwise definable, but CH fails. – Asaf Karagila Apr 10 '18 at 15:59
  • That's a good point. Do you know of any expository articles that try to look into the heuristic reasons why CH is independent? Although we could just say that it is a 'brute fact', this would be very unsatisfactory. When we look at something like Godel's incompleteness theorem, we can (kind of) give an intuition behind it since it would be naive to suppose that a single list of axioms can fully determine the colossal mathematical universe they generate, but when we have the specific example of CH, it is much harder to give a heuristic reason for it's independence. – Elie Bergman Apr 10 '18 at 16:06
  • Be careful of this "there are only countably many definitions" argument; there are subtle issues and it probably doesn't imply what you think it implies. In particular, it is consistent with ZFC that every set can be defined, so there need not be any "deep" reals at all! In my opinion, this is very similar to Skolem's paradox, in that the term "countable" is being used in two different senses. –  Apr 16 '18 at 19:14

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I think that what you're missing here—indeed, what many people miss—is that the Continuum Hypothesis is not a statement about the reals. It is a statement about the power set of the reals.

To see why, note that you can always force CH to hold without adding any reals to your model.1 But you cannot change the truth value of CH without adding new subsets of $\Bbb R$. If we translate this to the von Neumann hierarchy, CH is not a statement about $V_{\omega+1}$ (which is practically $\Bbb R$), but rather about $V_{\omega+2}$.

Now, let's talk about the intuition here. Let's fix a model of $\sf ZFC$, call it $V$. The von Neumann hierarchy tells you that the universe is generated by simply at each step taking all the subsets available to you.

The constructible hierarchy, on the other hand, lets you limit what you're adding to your universe, by letting you only add things that you can define. This lets you have a much finer structure to the universe, it lets you refine its construction, and thus provides you with a way to prove things you cannot prove about an arbitrary model of $\sf ZFC$.

This is not about undefinability of real numbers. This is not about undefinability at all. This is about sets of real numbers. And the reason that we cannot really prove anything about CH from $\sf ZFC$ itself is that the power set operation is "too wild" for $\sf ZFC$ to control with only first-order logic statements.2

So whereas $\Bbb N$ is something that $\sf ZFC$ has a firm and tight grasp over, its power set is a much looser object, and its power set is well outside the grasp of $\sf ZFC$.

This may or may not answer your question. And you might read this and exasperate in annoyance, that this is not what you signed up for, you're not even supposed to be working today. But that's set theory for you, indeed that's mathematics for you. Some ideas cannot be communicated "intuitively", unless you already have the right intuition to begin with. And while this is unfortunate, I can only comfort you by reminding you that beer is still a tangible object that not even $\sf ZFC$ can ruin for you.

Footnotes.

  1. With the usual caveat that we're talking about countable models. But that's not the important part here.

  2. You could argue that second-order $\sf ZFC$ is better here. But this is bit trickier. Any model of second-order $\sf ZFC$ is of the form $V_\kappa$ for some inaccessible $\kappa$, so it must agree with $V$ on the truth value of CH. This means that we delegate the question about CH to the meta-theory, to $V$, and then the question reverts to its original form.

Asaf Karagila
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    +1, although it isn't always clear to me how much control ZFC has over $\mathbb{N}$ even (cf. Hamkins' well-foundedness mirage). – Carl Mummert Apr 10 '18 at 16:12
  • I really like this answer! The crux of the issue is (as you have pointed out) that the power set function has a certain 'indeterminanteness' about it (i.e. is non-absolute). The question is, why does this particular axiom (power set) yield this property? I was guessing (but after your comments, I think this cannot be right) that the reason for this indeterminancy was to do with the fact that there are sets that cannot be defined in first order logic. This would open up the possibility of forcing the number of these sets to be various cardinals that we choose. – Elie Bergman Apr 10 '18 at 16:12
  • @Carl: Yes, but in the sense that it defines it in a fairly robust way that is absolute between transitive models. – Asaf Karagila Apr 10 '18 at 16:13
  • @AsafKaragila: I wonder whether that says more about transitive models (and the fact that we understand them much better than general models of ZFC) than it says about ZFC itself, though. – Eric Wofsey Apr 10 '18 at 16:14
  • @Eric: Well, we understand $\Bbb N$ better than we understand non-standard models of PA; and we understand $\Bbb R$ better than we understand the hyperreals or other non-Archimedean fields. I don't see why set theory has to be the odd-duck out. – Asaf Karagila Apr 10 '18 at 16:17