1

The title pretty much explains my question. I've been reading into basic set theory the past couple days and I was wondering if in ZFC we can prove the existence of a countable set that is non-constructible.

korn55
  • 313
  • Why would you want that? – Just dropped in Apr 29 '21 at 16:57
  • 9
    Constructible in the sense of Godel's $L$? We can't prove there is a non-constructible set period (it is consistent that all sets are constructible since $L$ is a model of ZFC in which all sets are constructible), so you'll have to be more specific about what you want. (Edit: I see you used the "constructive mathematics" tag, so maybe not, but in any event, you need to be more specific about what you want.) – spaceisdarkgreen Apr 29 '21 at 16:59
  • 2
    There are certainly subsets of $\mathbb N$ which are not recursively enumerable. – Thomas Andrews Apr 29 '21 at 17:05
  • @ThomasAndrews I don't see how you would go about proving that. Oh never mid, I think it's just a diagonal argument.. – korn55 Apr 29 '21 at 17:16
  • Isn't that trivial to prove? There are uncountably many subsets of $\mathbb N$ but only countably many can be recursively enumerable. What am I missing? – silver Apr 29 '21 at 17:18
  • 3
    @silver Why are you interpreting "constructible" as "recursively enumerable?" – Noah Schweber Apr 29 '21 at 17:22
  • @silver you aren’t missing anything (on the other hand I’m not sure what the basis would be for equating “constructible” with r.e.) – spaceisdarkgreen Apr 29 '21 at 17:25
  • @Noah I think that one is Thomas Andrews’s doing, not silver’s, but maybe I’m misreading the dynamics of the thread. – spaceisdarkgreen Apr 29 '21 at 17:27
  • @spaceisdarkgreen Fair point. – Noah Schweber Apr 29 '21 at 17:28
  • Right, I certainly didn’t mean my recursively enumerable example to be an answer, just an example from another definition of “constructible.” – Thomas Andrews Apr 29 '21 at 18:26
  • If there are any countable non-constructible sets, there are many. One well-known example: $0$#. I assume by "constructible" you mean... constructible, a member of $L$. Unless ZFC is inconsistent, it can't prove that any non-constructible sets exist: otherwise it could prove $V \neq L$, whereas ZFC + $V = L$ is consistent. – BrianO Apr 29 '21 at 21:17

1 Answers1

5

It depends what you mean by "constructible."

In set theory, constructibility has a very specific technical meaning according to which it is consistent with $\mathsf{ZFC}$ that every set (countable or not) is constructible, assuming of course that $\mathsf{ZFC}$ itself is consistent in the first place.

However, this notion of constructibility has nothing to do with constructive logic, nor with any sort of "computational" notion of construction. It's certainly the case that $\mathsf{ZFC}$ proves the existence of high-complexity (e.g. not computably enumerable) sets, but we shouldn't use the term "constructible" here.

Noah Schweber
  • 245,398
  • On the other hand, sometimes “constructible” is intuitively interpreted as “definable” and that’s another can of worms. – spaceisdarkgreen Apr 29 '21 at 17:32
  • @spaceisdarkgreen Indeed (for those interested see e.g. this paper or various discussions at MSE/MO such as here). – Noah Schweber Apr 29 '21 at 17:34
  • This sounds interesting, and I love mathematics. Are these topics that I could feasibly learn and appreciate at an introductory level, without needing to become an expert in set theory? And if so, are there any good books that you would recommend to someone with a master's in math, but who has taken mostly analysis courses (not graduate courses in logic, set theory, etc.)? – Joe Apr 29 '21 at 17:42
  • @Joe It certainly takes some work. You need a firm grasp of ZFC, the cumulative hierarchy, and definition by transfinite recursion as well as some first-order model theory. That will be enough to get you the definition of $L$, but then doing anything with it will require substantially more work. The key fact about $L$ is that it satisfies $\mathsf{ZFC+CH}$ (actually $\mathsf{GCH}$, and indeed much more, but this is the common starting point), which gives you one half of the independence of $\mathsf{CH}$ from $\mathsf{ZFC}$ (namely its un-disprovability). Forcing does the other half. – Noah Schweber Apr 29 '21 at 20:23
  • I'd say that nailing down all the details is about the equivalent of 1.5 graduate semesters? But this really varies substantially by person and presentation. – Noah Schweber Apr 29 '21 at 20:24
  • I wouldn’t say that I want to be able to do anything with it, just understand the basics, for my own personal fulfillment. I’ll look up books on the topics you’ve listed. If you have any favorites, please let me know. – Joe Apr 29 '21 at 21:33
  • @Joe You might try the first volume of Just and Weese’s Discovering modern set theory. It’s ending point is a section on $L$. I didn’t learn from it myself, but it seems user-friendly, though the style is chatty and some people like that more than others. The analysis of $L$ has a lot of technical details that are plausible but tedious to prove explicitly (similar in spirit to the incompleteness theorem)... they outright skip a lot of this stuff. More seriously they haven’t yet developed the necessary model theory results, so those are forward-referenced to the next volume. Still, not bad. – spaceisdarkgreen Apr 30 '21 at 02:24
  • @spaceisdarkgreen Thank you for the recommendation! I’ll try that book next. But I started “Set Theory and the Continuum Problem” by Smullyman & Fittting last night. I am a little concerned that they state in the intro their treatment of several topics is non-standard. Chapter one (general background) was very easy to follow though, so I’ll see how this book goes. Chapter 12 is when they get to “the class L of constructible sets”, and Chapter 16 introduces forcing. – Joe Apr 30 '21 at 12:16