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I am reading up on NBG set theory, a bit and am doing it fairly scarcely, as in I don't have a proper book on it and just check up on things I can find but I have a question. We have the class $V$ of all sets there, but what is the logic notation for it?

I know the infinite set can be written as

$\exists S(\exists x(x\in S\land\forall y(y\notin x)\land\forall z(z\in S\rightarrow\exists u(u\in S\land\forall w(w\in u\leftrightarrow u\in z\lor u=z))))$

while avoiding the empty set, but what is it for $V$? I feel it should be along the way of

$$\exists V:\emptyset\in V\land(x\in V\implies x \cup \{x\}\in V )\land S\in V \land (x\in V\implies\mathcal{P}(x)\in V)$$

I am not perfect in logical notation but to me it reads that the empty set is there (I am being lazy, we can reformulate to avoid it), then the "successor set" is also in there, to include all finite sets, then we include the previously defined infinite set, and from there all powersets are included as well. However I feel this is not entirely correct as I feel the continuum hypothesis is slipping in what I made. So what would be the formal definition exactly?

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You are confusing axioms with definitions.

The Axiom of Infinity does not define "the infinite set", because there are many infinite sets. It posits the existence of a set which we call "infinite", and it does so in the simplest way possible in the language of set theory (to some extent).

But what it really means, is that there is a set with the empty set as a member which is closed under the successor operation. But now we can prove there are in fact many such sets. In fact a proper class of these sets. So the Axiom of Infinity does not define "the infinite set". It does not even tell you an equivalent condition for a set to be infinite. It just tells you about the existence of a set which has the property that it is infinite.

The class $V$ of all sets, is the class of all the sets. In $\sf ZFC$ this just means all the objects, but indeed in $\sf NBG$ this means all the objects which are sets. Now recall that being a set in $\sf NBG$ means that you are an element of another class.

So $V$ is defined as $\{x\mid\exists y(x\in y)\}$. Using class comprehension, easily this class exists.

In what you wrote we have some problems:

  1. $S$ is not a fixed set in our language. Even worse, in $\sf NBG$, asserting existence does not mean that a set exists, but rather that a class exists. So you might not even get an infinite set.

  2. Even if you do mean that $S$ is a set whose existence is guaranteed by the Axiom of Infinity, there can be many such sets, as mentioned, and you just take a closure under power sets now. So just take $\{\mathcal P^n(S)\mid n\in\omega\}\cup\{\mathcal P^n(\varnothing)\mid n\in\omega\}$, and that would be a set satisfying these properties, but certainly this is a countable set.

    You seemed to try and capture the essence of the von Neumann hierarchy, which is a good intuition, but you need to require more. You need to require closure under unions, but here you run into a problem, that you want those to be set unions, and not class unions. This can be formalized, and the end result would looks a bit like Grothendieck's Universe definition. But nevertheless, it will invariably rely on the fact I mentioned earlier: sets are just classes which are members of other classes. So that's a simpler way to do it.

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  • $\begingroup$ Thank you, clear and nice! $\endgroup$ – Zelos Malum Mar 25 '17 at 8:31

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