I'm trying to study K. Kunen's "Set theory: introduction to independence proofs". Here's the following theorem which he gives to show that we can prove the existence of an infinite cardinal $>\omega$ in ZF (using the powerset):
I can't understand why the last sentence is true. Literally, why $\sup(S)$ is a cardinal and why it's $>\alpha$? What I could miss to make it really obvious, or it's just a construction that needs further verification?
Note that $S$ is a set of ordinals, and if $\beta\in S$, then there is a bijection between $\alpha$ and $\beta$. But also, if $\beta$ is an ordinal, and there is a bijection between $\alpha$ and $\beta$, then using this bijection we can define $R$ which well-orders $\alpha$ and witnesses that $\beta\in S$.
Now, $\sup(S)$ is some ordinal, $\kappa$. If there is a bijection between $\kappa$ and $\alpha$, then $\kappa\in S$, and therefore $\kappa=\max(S)$. But because $\alpha$ is infinite, so must be $\kappa$, and therefore $\kappa+1$ is also in bijection with $\kappa$ and $\alpha$. So $\kappa$ must be a greater in cardinality than any member of $S$, which means that it is a cardinal, and since $\alpha\in S$, $\kappa$ is a greater cardinal.
In addition, we can see that $\kappa$ is the smallest cardinal greater than $\alpha$, so we denote it often by $\alpha^+$, the successor cardinal of $\alpha$.