Is the successor of a non finite ordinal a non finite ordinal?

Question

Definition:

$$\alpha \text{ is finite iff } \forall\: \beta \: \text{ordinal}, \: \beta \leq \alpha\: \text{and}\: \beta \: \neq \emptyset \: \Rightarrow \exists \gamma( \beta=\gamma\cup\{\gamma\}) $$

(definition corrected thanks to suggestion)

Question: Does this hold true ?

$$\alpha \ \ \text{non finite} \Rightarrow \: \alpha + 1 \ \ \text{non finite} $$

If it is true, then a non finite ordinal can have a predecessor but a limit ordinal cannot have a predecessor, am I right ?

Answer 1

Your definition is slightly wrong. It merely states that every smaller ordinal is $0$ or a successor, but this much is also true for $\omega$, which is certainly not finite, what with it being the least infinite ordinal.

You can fix this by changing $\forall\beta<\alpha$ to $\forall\beta\leq\alpha$.

Now, to your question, of course that the successor of an infinite ordinal is infinite. If $\alpha$ is infinite there is a witness for the fact that $\alpha$ is not finite, which is either $\alpha$ or some $\beta<\alpha$. This will remain a witness when we consider $\alpha+1$.

Finally, a limit ordinal is defined as an ordinal which is not a successor.¹ So again, quite trivially, limit ordinals are not successors.

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1. In some places $0$ is also separated from the limit ordinals, in which case a limit ordinal is a non-zero ordinal which is not a successor.