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In the set theory of NBG, the class existence theorem says that for all predicative well formed formulas $\varphi$ (wff's in which the variables quantify only over sets) there is a (unique up to extensionality) class, $A$, such that if $x_1, x_2, ..., x_n$ are the free variables of $\varphi$ then the universal closure of $(x_1, x_2, ..., x_n) \in A \iff \varphi(x_1, x_2, ..., x_n)$ is true.

So in some sense we "can" define a function that takes as a parameter a predicative wff, let's call it $\varphi$ and outputs the unique class such that the universal closure of $(x_1, x_2, ..., x_n) \in A \iff \varphi(x_1, x_2, ..., x_n)$ is true.
But we can sort of define an inverse function, call it $g$. For each class $A$ define $g(A)= "x \in A"$. And now one can see that $g$ is "injective". So in some sense, the cardinality of the whole classes is less that that of the predicative wff's, which the latter is clearly at most countable. And this result looks like a contradiction, like wouldn't it imply that the Von Neumann universe is at most countable?

I feel like there is a huge flaw in my reasoning such that I am making my function $g$ that operates on objects of the language($A$, the classes) and outputs objects in the metalanguage(predicative wff's).

Can someone explain the flaw in this reasoning? I possess a limited understanding of formal logic and the axiomatic set theory of NBG, without reaching an advanced level of expertise.

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The formula $x\in A$ uses the class $A$ as a parameter. There are only countably many formulas, but if we allow arbitrary classes to appear as parameters in these formulas, then there are at least as many formulas with parameters as there are classes!

Alex Kruckman
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  • Yes, but if I have an injection from $M$ to $N$ then the cardinality of $M$ is at most the cardinality of $N$. And my function $g$ is injective due to extensionality. That was my original issue to begin with. – Shthephathord23 Feb 02 '24 at 15:21
  • @Shthephathord23 Aside from your function is not well-defined, you did not consider the case when $A$ is not a predicative class. – Hanul Jeon Feb 02 '24 at 15:35
  • @Shthephathord23 I'm not sure you understood the point of my answer. $g$ is not an injective function from classes to formulas (of which there are countably many), it is a function from classes to formulas with parameters (and there is no reason to think that there are only countably many of the latter). – Alex Kruckman Feb 02 '24 at 16:27
  • But aren't formulas with parameter a subset of the set of formulas? Since each is a formula? – Shthephathord23 Feb 02 '24 at 17:13
  • @Shthephathord23 A formula with parameters is a formula in which elements of some structure have been substituted for some of the free variables. For example, although there are only countably many formulas in the language of rings, there are continuum-many formulas in the language of rings when we allow parameters from $\mathbb{R}$, since, for example, $x = r$ is a formula with parameters from $\mathbb{R}$ for each real number $r$. – Alex Kruckman Feb 02 '24 at 18:59
  • okay, I think I understand now, but it seems a bit counterintuitive that the set/class of formulas with parameters are not a subset of the whole set of formulas...If I would go by intuition alone, I would say that formulas with parameters are a proper subset/subclass of the whole collection of formulas. But that leads to a contradiction... Maybe what I am asking is stupid, but may you explain this in a way that is intuitive? Or maybe provide me with some resources so i can research this topic? I did not find anything related on the internet. – Shthephathord23 Feb 02 '24 at 19:18
  • Like what I would be looking in for an answer is the reason why formulas with parameters are not a subset of the whole formulas. – Shthephathord23 Feb 02 '24 at 19:20
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    @Shthephathord23 Well, this is why mathematics is about precise definitions, not intuition. Does this help? https://math.stackexchange.com/q/4251553/7062 – Alex Kruckman Feb 02 '24 at 20:12
  • @Shthephathord23 Maybe it would be clearer if I said this: why are there only countably many formulas? Because each formula is a finite string from a countable alphabet containing symbols like $\in$, $=$, $\land$, $x$, $y$, $z$, etc. Your "formula" $g(A)$ is $x\in A$, but here $A$ is a class, which is not one of the symbols in our countable alphabet! – Alex Kruckman Feb 02 '24 at 20:21
  • The link that you posted definitely helped. If you happen to know any good resources on learning formal logic, or maybe recommend/share them here, I would gladly appreciate. Thanks! – Shthephathord23 Feb 02 '24 at 20:32