TITLE: Question on a proof from Jech's Set Theory of: If $X$ is a set of cardinals, then $\sup X$ is a cardinal.
QUESTION [7 upvotes]: The proof: Let $\alpha= \sup X$. If $f$ is a bijective mapping of $\alpha$ onto some $\beta < \alpha$, let $\kappa \in X$ be such that $\beta < \kappa \le \alpha$. Then $|\kappa|=|\{f(\xi): \xi < \kappa\}| \le \beta$, a contradiction. Thus, $\alpha$ is a cardinal.
*An ordinal $\alpha$ is a cardinal if $|\alpha|\neq |\beta|$ for all $\beta < \alpha$.

Here's how far I think I understand the logic of the proof; first we assume $\alpha$ isn't a cardinal. This means there would be a $\beta$ less than $\alpha$ which is equinumerous to $\alpha$. We then suppose another ordinal $\kappa$ greater than $\beta$ but that injects into $\alpha$. What I don't get is how do we know that $|\kappa|=|\{f(\xi): \xi < \kappa\}|$, not to mention how $|\{f(\xi): \xi < \kappa\}| \le \beta$. If all of that stands true, then I get that we arrive at the contradiction; $\beta < \kappa$ means $\beta \in \kappa$ and that implies there can't be an injection from $\kappa$ into $\beta$ ($\kappa \le \beta$).

REPLY [3 votes]: Let $X$ be a set of cardinals. Let $\alpha = \sup X$. Suppose that $\sup X$ is not a cardinal. Then there exists a $\beta < \alpha$ such that there exists a bijection $f : \alpha \rightarrow \beta$. Since $\alpha$ is not a cardinal and $X$ is a set of cardinals, you have that $\alpha \notin X$. Since $\alpha = \sup X$, you must have that there exists a cardinal $\gamma \in X$ such that $\beta < \gamma < \alpha$. Thus $f|\gamma$ ($f$ restricted to $\gamma$) is a injection from $\gamma$ to $\beta$. By definition of $\beta < \gamma$ (as ordinals), you have that injection from $\beta$ into $\gamma$. By the Shoder-Bernstein Theorem, there is a bijection between $\gamma$ and $\beta$. But $\gamma \in X$ is a cardinal; hence, there can not be a bijection between $\gamma$ and the smaller ordinal $\beta$. Contradiction!