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Working in a consistent theory (say, ZFC), is there a set/class that is provably nonempty, but nothing is provably in it?

Formally, ($T\vdash \exists x:x\in A)\land (\forall x,T\nvdash x\in A)$

Fei
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    If you can prove $\exists x(x\in Y)$, then something is provably in $Y$; you may have to be more precise about what you're looking for. – Malice Vidrine Dec 04 '18 at 21:12
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    Your question needs some clarification: (1) presumably your theory is a theory in a language suitable for set theory (i.e., at the very least it contains a predicate symbol for the membership relation). (2) presumably by "nothing is provably in it", you mean that we can't exhibit a term of the language which belongs to the set. What constants and function symbols does your language contain? (Set theory is often formalised without any constants or function symbols.) – Rob Arthan Dec 04 '18 at 21:16
  • Thanks for your comments – Fei Dec 04 '18 at 21:39
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    You haven't clarified Rob Arthan's second point at all. What does "$\forall x,T\nvdash x\in A$" mean? What can $x$ be here? – Eric Wofsey Dec 04 '18 at 21:46
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    A reasonable interpretation of the question would be to say that $x$ and $A$ really are definitions of sets in ZFC. In other words, there is a formula $\varphi(y)$ and ZFC proves there is a unique $A$ such that $\varphi(A)$ is true, and also that this unique $A$ is nonempty. Then the question would be whether there is a formula $\psi(z)$ such that ZFC proves there is a unique $x$ such that $\psi(x)$ is true, and such that ZFC proves this $x$ is in $A$. But with this interpretation, the answer you have accepted is wrong. – Eric Wofsey Dec 04 '18 at 21:53
  • Why is it not trivial that $(\exists x:x\in A)\implies x\in A$? If we can write $x$ in the first case then surely it can be written in the implication? – it's a hire car baby Dec 04 '18 at 22:18
  • @RobertFrost "$x\in A$" isn't a sentence; it doesn't make sense to ask about its truth value. A concrete example: would you consider "There is an $x$ which is even, so $x$ is even" to be valid reasoning? – Noah Schweber Dec 04 '18 at 22:19
  • @NoahSchweber I get you, thanks. – it's a hire car baby Dec 04 '18 at 22:25
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    The question is too imprecise as stated. For instance, there are no terms in the language of ZFC, but your formulation suggests otherwise. – Andrés E. Caicedo Dec 05 '18 at 00:53

2 Answers2

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In ZFC, the set of ordinals $\alpha$ for which $\aleph_\alpha$ is the cardinality of $\Bbb R$ has one element, but you can't determine what it is. It could be any non-zero finite ordinal, for example.

J.G.
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    But "the unique $\alpha$ such that $|\mathbb{R}|=\aleph_\alpha$" is a perfectly good definition of a set in ZFC. Intuitively this may seem like cheating, but it's very difficult to come up with a clear reason why such a definition would not be allowed but other ways of defining specific ordinals would be. – Eric Wofsey Dec 04 '18 at 21:49
  • Your answer is assuming that the logical language for ZFC includes function symbols such that you can write down terms denoting $\Bbb{R}$ and its cardinal. To answer the question you need to be clear about what the language is and what definitional principles are allowed (cf. Eric's comment). – Rob Arthan Dec 04 '18 at 21:50
  • @I'm allowed to define predicates to abbreviate things. In particular, first I can define what a function is, what a bijection is and what it means for one set to be equipollent to another, and then I can define alephs in terms of ordinals' Hartogs numbers. – J.G. Dec 04 '18 at 21:55
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    I am not sure this quite works. You can define $\alpha$ as the unique ordinal such that $\aleph_\alpha=|\mathbb R|$. Your theory proves that this $\alpha$ is in your set. Thus is kind of unsatisfactory, but without more concrete restrictions in the formulation of the question, this shows your example does not work. – Andrés E. Caicedo Dec 05 '18 at 00:30
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    (If you want to exclude such definitions, then you need to explain why this is allowed for your set but not for its element.) – Andrés E. Caicedo Dec 05 '18 at 01:00
  • I wonder if my question makes more sense if I say "working in a model of T" instead of "working in T" (so x really means "x in [the universe of] M"). If so I guess that's probably what I meant initially. In this case I think J.G.'s answer is correct with the model being V (A would be a class in V). – Fei Dec 06 '18 at 22:20
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Here's a possible interpretation of your question, and a possible answer.

Is there a formula $\varphi(x)$ in the language of set theory (with no parameters) such that:

  • $\text{ZFC}\vdash \exists x\, \varphi(x)$.
  • For every formula $\psi(y)$ in the language of set theory (with no parameters), if $\text{ZFC}$ proves that there is a unique set $Y$ satisfying $\psi(Y)$, then $\text{ZFC}$ does not prove that $Y$ satisfies $\varphi(x)$. That is, if $\text{ZFC}\vdash \exists^! y\, \psi(y)$, then $\text{ZFC}\not\vdash \forall y\, (\psi(y)\rightarrow \varphi(y))$.

The gloss is: Is there a class which is provably nonempty, but such that there is no definable object which is provably an element of the class?

I believe (maybe someone can provide a proof or correct me if I'm mistaken) that taking $\varphi(x)$ to say "$x$ is a well-ordering of the reals" satisfies these criteria.

You can also give a version about sets, rather than proper classes.

Is there a formula $\varphi(x)$ in the language of set theory (with no parameters) such that:

  • $\text{ZFC}$ proves that there is a unique set $X$ satisfying $\varphi(x)$.
  • $\text{ZFC}$ proves that $X$ is non-empty.
  • For every formula $\psi(y)$ in the language of set theory (with no parameters), if $\text{ZFC}$ proves that there is a unique set $Y$ satisfying $\psi(y)$, then $\text{ZFC}$ does not prove $Y\in X$.

Here we can take $\varphi(x)$ to say "$x$ is the set of all well-orderings of the reals".

The example in J.G.'s answer does not satisfy these criteria (and this is what the comments from Eric and Andrés are about). There, the formula $\phi(x)$ says "$x$ is the set of ordinals $\alpha$ such that $|\mathbb{R}| = \aleph_\alpha$." Then $\text{ZFC}$ proves that there is a unique set $X$ satisfying $\varphi(x)$, and $X = \{\alpha\}$ is a singleton. But now if we take the formula $\psi(y)$ to say "$|\mathbb{R}| = \aleph_y$", then $\text{ZFC}$ proves that there is a unique $\alpha$ satisfying $\psi(\alpha)$, and $\alpha\in X$.

Alex Kruckman
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  • Thanks Alex, I think I understand your (& Rob, Eric and Andrés') point now. The interpretation you gave is a very interesting one, and perhaps the most legitimate given the lack of clarification in the question. Besides, I think "well-orderings of reals" is a perfect answer to this interpretation. Great answer. – Fei Dec 06 '18 at 21:53
  • Great! Just in case you don't know (since I see you're a new user of this site) you can upvote answers you find helpful by clicking the up arrow by the number on the left. – Alex Kruckman Dec 06 '18 at 21:57