Is the class of ordinals part of the class of sets in NBG set theory?

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In Von Neumann–Bernays–Gödel set theory (NBG), the fundamental objects are classes, not sets. But, in addition to classes, in NBG we have some sets. I took for granted that in NBG we have the following two clases:

  1. The class $\mathcal{C}_1$ of all ordinary sets
  2. The class $\mathcal{C}_2$ of all ordinals

It seems intuitively that $\mathcal{C}_2$ is part of $\mathcal{C}_1$. In NBG, a set is defined to be a class that is an element of some other class, because of this it is impossible that $\mathcal{C}_2 \in \mathcal{C}_1$, because this would entail that $\mathcal{C}_2$ is a set, but it is not the case.

My questions are:

  1. Can we in NBG write $\mathcal{C}_2 \subset \mathcal{C}_1$ in some meaningful sense?
  2. What prevents in NBG to construct explicitly a class $\mathcal{C}_3$ such that $\mathcal{C}_2\in\mathcal{C}_3$ or $\mathcal{C}_1\in\mathcal{C}_3$?
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To the first part of your Question, yes, it does make sense to define subclass relations between classes, e.g. that $\mathcal{C}_2 \subset \mathcal{C}_1$ because $x\in \mathcal{C}_2 \implies x\in \mathcal{C}_1$ for all sets $x$.

The second part of your Question needs clarification in that "[w]hat prevents" a proper class (a class that is not a set) from being said to be a member of another class is that by definition the members of classes are always sets. So to do what you describe would require constructing classes $\mathcal{C}_1$ or $\mathcal{C}_2$ as sets.

So perhaps you want to know what prevents us from constructing the class of all sets as a set, etc. However I suspect that might be well-known to you, as the point of the axioms of NBG are to avoid (so far as we know) the classic paradoxes of naive set theory.