5
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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? $\endgroup$ – Asaf Karagila Apr 10 '18 at 13:24
  • $\begingroup$ I don't think so. The Cantor diagonal series is exactly defined. $\endgroup$ – Rudi_Birnbaum Apr 10 '18 at 13:25
  • $\begingroup$ 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. $\endgroup$ – user247327 Apr 10 '18 at 13:25
  • $\begingroup$ 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... $\endgroup$ – Isky Mathews Apr 10 '18 at 14:38
  • 2
    $\begingroup$ 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. $\endgroup$ – Asaf Karagila Apr 10 '18 at 15:59
6
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ +1, although it isn't always clear to me how much control ZFC has over $\mathbb{N}$ even (cf. Hamkins' well-foundedness mirage). $\endgroup$ – Carl Mummert Apr 10 '18 at 16:12
  • 3
    $\begingroup$ @realdonaldtrump: Do you have a better alternative from a foundational point of view? Because anything stronger than FOL is going to suck at not having a completeness theorem to its nature, or worse; and everything weaker than ZFC as a foundational theory would have even less powers dealing with TOL. $\endgroup$ – Asaf Karagila Apr 10 '18 at 16:14
  • 2
    $\begingroup$ @realdonaldtrump: Then you also don't see the need to have an algorithmic way to determine if a proof is valid or not? That's an interesting approach to mathematics. $\endgroup$ – Asaf Karagila Apr 10 '18 at 16:25
  • 1
    $\begingroup$ @realdonaldtrump: So what? You can't even determine if a group is abelian. What's your point? $\endgroup$ – Asaf Karagila Apr 10 '18 at 16:31
  • 2
    $\begingroup$ @realdonaldtrump: No? I'm just not sure what you're getting it. Are you trying to say that it's wrong to celebrate an incompleteness foundations of mathematics? That's one opinion. But I hold another. And it seems that right now you're trying to argue that your opinion is better than mine. So I'm a bit confused as to what do you think you're going to achieve with this, or where do you think you're going. But then again, why do I expect from someone who claims to be the real Donald Trump... I think I'll get back to writing my papers. $\endgroup$ – Asaf Karagila Apr 10 '18 at 16:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.