For every ordinal $\alpha$, there is a cardinal number greater than $\alpha$ without AC

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Let $\alpha \geq \omega$ be an ordinal number. Let $\operatorname{type}(A, R)$ be the only ordinal number $\beta \simeq (A, R)$. We define :

$$W = \{R \in \mathcal{P}(\alpha \times \alpha) : R \text{ is a well order} \}$$

$$S = \{ \operatorname{type}(\alpha, R) : R \in W \}$$

Then, how to prove $\sup(S) = \bigcup S$ is a cardinal number greater than $\alpha$ ?

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First note that $\in$ is a well ordering of $\alpha$, so it is in $W$; then assume towards contradiction that $\sup(S)$ is not a cardinal, this means that is equipotent to some smaller ordinal. Hit by definition all the ordinals below $\sup(S)$ are at most of size $\alpha$, and from here infer a contradiction.

Note, also, that the definition as you write it only works for infinite ordinals, which is fine since we can prove this fact by hand for finite ordinals. But you can modify the definitions so this theorem will be uniformly applicable to all ordinals.