Question regarding the proof of the Hartogs cardinal theorem.

Question

I am reading through the text The Foundations of Mathematics by Kenneth Kunen. On pages 54 and 55 he gives the following proof of the Hartogs cardinal theorem.

Theorem For every set $A$, there is a cardinal $\kappa$ such that $\kappa \npreceq A$.

Proof: Let $W$ be the set of pairs $(X,R) \in \mathcal{P}(A) \times \mathcal{P}(A \times A)$ such that $R \subseteq X \times X$ and $R$ well-orders $X$. So, $W$ is the set of all well-orderings of all subsets of $A$. Observe that $\alpha \preceq X$ iff $\alpha = \text{type}(X,R)$ for some $(X,R) \in W$ (See Exercise I.$11.19$). Applying the Replacement Axiom, we can set $\beta = \text{sup}\{\text{type}(X,R) + 1: (X,R) \in W\}$. Then $\beta > \alpha$ whenever $\alpha \preceq A$, so $\beta \npreceq A$. Let $\kappa = |\beta|$. Then $\kappa \approx \beta$, so $\kappa \npreceq A$. $\Box$

The exercise referred to in the proof is that for any set $A$, it can be well-ordered in type $\alpha \in \text{On}$ iff there is a bijection between them, so $\alpha \approx A$. I completely understand every other part of the proof except the following. So the third line in the proof seems to be using a modified version of this statement. Clearly if $\alpha = \text{type}(X,R)$ for some $(X,R) \in W$, then by the exercise there is a bijection $f: \alpha \rightarrow X$, which will also be an injection so that $\alpha \preceq X$. My issue is with going the other way. Suppose now that $\alpha \preceq X$. Then for $\alpha$ to be $\text{type}(X,R)$ for some well ordering $R$ of $X$, we would need first of all an order isomorphism of $(\alpha, \in)$ with $(X,R)$ which would need to be a bijective function, but how can I be guaranteed the existence of a bijective function with just an injection of $\alpha$ into $X$ given?

Answer 1

This is just a typo. In the sentence

Observe that $\alpha \preceq X$ iff $\alpha = \text{type}(X,R)$ for some $(X,R) \in W$ (See Exercise I.$11.19$).

it should say $\alpha\preceq A$ instead of $\alpha\preceq X$. (It does not even make sense to say $\alpha\preceq X$, since no specific $X$ has been defined and the $X$ appearing later in the sentence is a bound variable confined to the right side of the "iff".)

Specifically, if $\alpha\preceq A$ then there is an injection $f:\alpha\to A$, and then we can take $X$ to be the image of $f$ so $f$ gives a bijection between $\alpha$ and $X$. So, by the exercise, there exists a well-ordering $R$ of $X$ with order-type $\alpha$.