Given two classes $x$ and $y$, does this statement hold $x \in y \lor x \not \in y$? I know that if $x$ is a proper class, then it should hold since $x$ is a proper class if and only for every class $z$, then $x \not \in z$. However, if $x$ is a set, then intuitively it seems that this statement will hold, but I know that in mathematics there are some statements that cannot be proved using the current axioms. Sorry I am not too acquainted with ZFC or NBG.

# Set Theory – Basic Question in Axiomatic Set Theory

elementary-set-theoryset-theory

#### Related Solutions

The usefulness of the hyperreals $\mathbb{R}^*$ stems from such tools as saturation and the transfer principle. These tools are available in other superfields R' of R only to the extent that one can construct morphisms between the hyperreals and such R'. Even if some limit ultrapower model of $\mathbb R^*$ is maximal and unique in some weak sense, this does not really affect its applications, may of which rely on the simplest model of the hyperreals, namely $\mathbb R^{\mathbb N}$ modulo an ultrafilter on $\mathbb N$. Ultimately the relationship to other superfields of $\mathbb{R}$ has little bearing on whether "Robinson's program has been completed successfully" as you put it.

This is because the framework described by the NBG axioms is strictly richer than that which the ZFC axioms describe. This is not an isolated phenomenon. Let me show you diffrent, perhaps a bit less abstract examples.

First example.

It is a well-known consequence of compactness that there is no first-order axiom in the language with just equality whose models are exactly the infinite sets. We need infinitely many axioms saying that there are at least $n$ elements, for arbitrarily large $n$.

On the other hand, if we add a linear ordering, we can consider the theory of dense linear orderings (with at least two points), which is finitely axiomatisable and has no finite models.

Every infinite set can be expanded to a dense linear ordering. Thus, the pure sets underlying dense linear orderings are exactly all infinite sets.

Second example.

For a field $F=(F,+,\cdot,0,1)$, the following are equivalent:

- $F$ is formally real, i.e. whenever a sum of squares is zero, all of them have to be zero.
- There is a linear ordering $\leq$ on $F$ which is compatible with the field operations (making $(F,+,\cdot,0,1,\leq)$ an ordered field).

Now, the first order of formally real fields is not finitely axiomatisable: for any $n$, there is a field which is not formally real, but such that the sum of at most $n$ squares, not all zero, is necessarily nonzero. The conclusion follows by compactness.

On the other hand, the theory of ordered fields is obviously finitely axiomatisable.

In this case, the pure fields described by the (finitely axiomatisable) theory of ordered fields are exactly the same as the pure fields described by the (not finitely axiomatisable) theory of formally real fields.

In both examples, the extra structure provided by the linear ordering allows us to force infinitely many axioms to hold simultaneously. Similarly, for NBG, the extra structure provided by the addition of proper classes as elements allows us to force infinitely many axioms to hold simultaneously.

In all cases, the cost is that we need to choose this extra structure: there are many ways to choose a dense linear ordering of an infinite sets, there are (typically) many ways to choose a compatible linear ordering for a formally real field, and there are also (typically?) many ways to expand a model of ZFC to a model of NBG - a pure set does not know its dense linear orderings, a formally real field (usually) does not know any particular compatible ordering, and there is no reason for a model of ZFC to know what classes you should endow it with to get a model of NBG.

## Best Answer

I have made my comment into an answer since the OP seemed happy with it.

The Law of Excluded Middle: In classical predicate logic (and some other types of logic), $\Psi \lor \lnot \Psi$ for all statements $\Psi$.$x\in y$ is a statement in the language of ZFC, and ZFC is a theory in predicate logic. Therefore $(x\in y)\lor \lnot (x\in y)$.

The intuitionist school of logic holds the following position:

It follows that in intuitionist logic we cannot use the law of excluded middle to prove statements.