Class of Successor ordinals.

Question

I understand fairly well why the class of all of the ordianls is not a set, because if we assume by contradiction that it is a set, then as an ordinal - we get that it contains itself - which it is a contradiction.

However, why is the class of all the successor ordinals isnt a set?

Answer 1

There are various ways to show this:

• Probably the most natural is to apply the replacement axiom (scheme): considering the formula "$x$ is the predecessor of $y$" we see that if the class of successor ordinals were a set, then the class of all ordinals would also be a set.

• We could "close downwards" via the axiom of union: can you show that if $S$ is a set of ordinals then $\bigcup S$ is the set of all ordinals smaller than some element of $S$? Now think about the downwards closure of the class of successor ordinals.

• We could consider the class of successor ordinals itself as a well-ordering with respect to "$\in$" - if it were a set it would have to be in bijection with some ordinal, but at the same time we can show that the class of successor ordinals is order-isomorphic to the class of ordinals (consider $\alpha+1\mapsto\alpha$).

• In the $\mathsf{ZF}$-style context, we can just observe that for every ordinal $\alpha$ there is a successor ordinal not in $V_\alpha$ - and so the class of successor ordinals isn't a set since every set lies in some $V_\alpha$.

Answer 2

Theorem: A class $C$ of ordinals is a set iff it has an upper bound.

Proof:

If $\alpha$ is an upper bound of $C$, then all elements of $C$ are elements of $\alpha+1$, as that is the set of all ordinals $\le\alpha$. Therefore $C=\{x\in\alpha+1: x\in C\}$, which by the axiom of separation is a set.

On the other hand, if $C$ is a set, then by the axiom of union, the union of its elements is a set, too. But if $C$ is unbounded, that union is the class of all ordinals, which is proper. Therefore $C$ has to be proper as well. $\square$

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Now it is easily seen that the class of successor ordinals is unbounded: Assuming that $\alpha$ is an upper bound, then $\alpha+1$ is a successor ordinal larger than the upper bound, contradicting the assumption.

Since the class of successor ordinals is unbounded, it is proper.