Question on a proof from Jech's Set Theory of: If $X$ is a set of cardinals, then $\sup X$ is a cardinal.

Acid2 10/14/2018 at 07:27. 3 answers, 0 views
set-theory cardinals ordinals

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 $\beta$ ($\kappa \le \beta$).

3 Answers

user642796 09/07/2012.

(I will write $X \approx Y$ to denote that $X$ and $Y$ are equinumerous.)

Note the following simple fact: If $f : X \to Y$ is a bijection, then for every $A \subseteq X$ the restriction $f \restriction A$ is a bijection between $A$ and $f''A$, which gives us that $A \approx f''A$ for all $A \subseteq X$.

In the context of the proof above, we have that $\kappa \approx f''\kappa = \{ f(\xi) : \xi \in \kappa \}$. As $f(\xi) \in \beta$ for all $\xi \in \alpha$, then surely $f''\kappa \subseteq \beta$. We thus have that $f''\kappa \subseteq \beta \subseteq \kappa$, and so by Cantor–Bernstein–Schroeder we may conclude that $f''\kappa \approx \beta \approx \kappa$. This contradicts the fact that $\kappa$ is a cardinal!

I would have proceeded slightly differently and avoided the troubling function $f$.

If $\alpha = \sup (X)$ is not a cardinal, then $|\alpha| < \alpha$ (where $|\alpha|$ denotes the unique cardinal equinumerous to $\alpha$). As $|\alpha|$ cannot be an upper bound for $X$, there is a cardinal $\kappa \in X$ such that $|\alpha| < \kappa \leq \alpha$. Note that by Cantor–Bernstein–Schroeder (as $|\alpha| \subseteq \kappa \subseteq \alpha$) it follows that $|\alpha| \approx \kappa$, contradicting that $\kappa$ is a cardinal!

William 10/31/2012.

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!

martini 09/07/2012.

As $\alpha$ isn't a cardinal and $\kappa \in X$ is one, we must have $\kappa < \alpha$. As $f$ is bijective, we have \[ \kappa = |\kappa| = |f''\kappa| = |\{f(\xi) \mid \xi < \kappa\}| \le |\{f(\xi) \mid \xi < \alpha \}| = |\beta| < \kappa. \] - Download Hi-Res Songs

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