Kunen's 10.16 - bigger cardinals without Axiom of Choice Kunen's Set Theory: An Introduction to Independence Proofs has Theorem 10.16, proving that for any cardinal there's a cardinal larger than it, without the Axiom of Choice (with AC it's easily done by Cantor's argument). The proof is light on the details and I'm trying to work them out - help?
Let $\kappa$ be an infinite cardinal, and let $W = \{R: R \textrm{ well-orders } \kappa\}$. $W$ exists by Power Set, being a subset of $P(\kappa \times \kappa)$. Let $S = \{\textrm{type}(\kappa, R) : R \in W\}$. S is a set of ordinals that exists by Replacement. Let $\alpha = \bigcup S$, then $\alpha$ is a cardinal greater than $\kappa$ (stated without proof).
Proof: $\alpha$ is an ordinal. Because $W$ includes $R=\in$, the "normal" well-order on $\kappa$, $S$ has $\kappa$ itself as a member. But in fact the ordinal $\kappa + 1$ (for instance) also well-orders $\kappa$, so $\kappa+1 \in S$, and $\kappa\in\kappa+1$ is in $\alpha=\bigcup S$. Therefore $\alpha$ is an ordinal strictly greater than $\kappa$.
It remains to prove that $\alpha$ is a cardinal, that is, there is no $\beta < \alpha$ in bijection with $\alpha$. Supposing there is, $\beta \in \alpha=\bigcup S$ means $\beta \in \gamma$ for some $\gamma \in S$, that is, an ordinal equivalent to a well-order on $\kappa$. I'm getting stuck somewhere around here, not sure how to get a contradiction (as an ordinal, $\beta$ could be quite larger than $\kappa$).
A different route: let's look at $|\alpha|$, which exists without AC because $\alpha$ is an ordinal. $|\alpha|$ is at least $\kappa$ because $\kappa$ is a member, thus a subset, of $\alpha$. Assume $|\alpha|=\kappa$ and let $f:\kappa\to\alpha$ be a bijection. This lets the $\in$-order on $\alpha$ well-order $\kappa$ and so we must have $\alpha \in S$. But then $\alpha+1$ is also in $S$, meaning $\alpha \in \alpha+1 \subset \bigcup S = \alpha$, meaning $\alpha \in \alpha$ which is impossible in ordinals even without Foundation, and that's the desired contradiction.
Did I do this right? I'm really uncertain about my proof, and I didn't even prove that $\alpha$ is a cardinal as the theorem states, only that its cardinal is greater than $\kappa$. Is there a better way?
Thanks!
 A: The theorem is actually a consequence of a more general statement about cardinalities. We can prove 10.16 as follows:

*

*We show that for every set $A$ there exists a minimal ordinal $\alpha$ such that there is no injection $\alpha\to A$

*We show that such an $\alpha$ is a cardinal number

*If $A=\kappa$ is a cardinal number, then we show that $\kappa<\alpha$
1
For any set $A$, there exists some ordinal $\alpha$ such that there exists no injective function $\alpha\to A$. The least such ordinal $\alpha$ is called the Hartogs number of $A$. The proof that every set has a Hartogs number is like the proof you stated.
Firstly, we can construct the set $W=\{R\subseteq A\times A\mid R\text{ is a well-order}\}$, then $W$ is a set by the Axioms of Power Set and Separation, and hence $S=\{\text{type}(R)\mid R\in W\}$ is a set of ordinals by the Axiom of Replacement, and thus $\bigcup S=\alpha$ is an ordinal. We claim that $\alpha$ is the Hartogs number of $A$:
Suppose that $f:\alpha\to A$ were an injection, then $\mathrm{ran}(f)$ is a well-ordered subset of $A$ under the ordering $f(x)\mathrel Rf(y)$ iff $x\leq y$, where $x,y\in \alpha$. But then $R$ well-orders a subset of $A$, so $R\in W$, and thus $\mathrm{type}(R)=\alpha\in S$, implying that $\alpha\in \bigcup S=\alpha$, which is impossible.
2
The Hartogs number $\alpha$ is a cardinal number: if not, then there exists some ordinal $\beta<\alpha$ equinumerous to $\alpha$, so we can formulate an injective function $h:\alpha\to\beta$. Since $\beta<\alpha$, by minimality of $\alpha$ as the Hartogs number there exists an injective function $f:\beta\to A$, but then $f\circ h:\alpha\to A$ would be an injective function as well, contradicting the definition of $\alpha$ as the Hartogs number of $A$.
3
If $A=\kappa$ is a cardinal number, then it is well-ordered, thus if we define $x\mathrel R y$ iff $x\leq y$, where $x,y\in \kappa$, then $R$ is a well-order isomorphic to $(\kappa,\in)$. Therefore $R\in W$, and $\mathrm{type}(R)=\kappa\in S$, showing that $\kappa<\bigcup S=\alpha$.

As a corollary to 3., it follows that if $\kappa<\beta<\alpha$, then $|\beta|=\kappa$: since $\beta<\alpha$ there is an injective function $\beta\to\kappa$, and since $\kappa<\beta$ there is an injective function $\kappa\to\beta$, thus by the Cantor-Bernstein-Schröder theorem $|\beta|=\kappa$.
Therefore, we can use Hartogs numbers to define the aleph numbers. Let $\aleph_0=\omega$; for any ordinal $\xi$ let $\aleph_{\xi+1}$ be the Hartogs number of $\aleph_\xi$; and if $\gamma$ is a limit ordinal, let $\aleph_\gamma=\bigcup_{\xi<\gamma}\aleph_\xi$.
A: Your definition of $\alpha$ isn't quite right. You have to define $\alpha = \{type(S, R) \mid R$ well-orders $S \subseteq \kappa\}$.
I claim that $x \in \alpha$ if and only if $x$ is an ordinal and there is some injection $x \to \kappa$.
For suppose $x \in \alpha$. Then take some $R$ which well-orders $S \subseteq \kappa$ such that $x = type(R)$. Take a bijection $f : x \to S$. Then $f : x \to \kappa$ is an injection.
Conversely, suppose that $x$ is an ordinal there is some injection $f : x \to \kappa$. Let $S$ be the range of $f$; then $f : x \to S$ is a bijection. So we can transfer the well-order on $x$ to a well-order $R$ on $S$, and thus $x = type(S, R)$.
From here, we can show that $\alpha$ is an ordinal. To do this, we note that $\alpha$ is a set of ordinals. And suppose $x < y \in \alpha$. Take $f$ such that $f : y \to \kappa$ is an injection; then $f : x \to \kappa$ is also an injection. Then $x \in \alpha$. Thus, $\alpha$ is a transitive set of ordinals, hence an ordinal.
Now, we show that $\alpha$ is a cardinal. For suppose there is some $\beta < \alpha$ and some bijection $f : \alpha \to \beta$. Take some injection $g : \beta \to \kappa$. Then $g \circ f : \alpha \to \kappa$ is an injection. Then since $\alpha$ is an ordinal and there is an injection $\alpha \to \kappa$, we have $\alpha \in \alpha$. That is, $\alpha < \alpha$. This is a contradiction.
Finally, I would note that you don't actually need anywhere near the full power of replacement for this proof. In fact, you only need Mostowski's Collapse Lemma to prove this statement. If we rephrase the theorem to discuss well-orders and initial well-orders instead of ordinals and cardinals, we don't even need Mostowski collapse to prove it.
