Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
The Cantor–Bernstein–Schroeder theorem (en.wikipedia.org/wiki/Cantor_Schroeder_Bernstein_Theorem) should help to construct the bijection you are looking for. –  Trevor Wilson Sep 7 '12 at 14:56

3 Answers 3

up vote 8 down vote accepted

(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!

share|improve this answer
    
Thank you for your help! –  Acid2 Sep 9 '12 at 5:25

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!

share|improve this answer
    
Thank you for your help! –  Acid2 Sep 9 '12 at 5:24

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

share|improve this answer
    
I see. But you're not explaining how, like I asked in the question, $|\kappa|=|\{f(\xi): \xi < \kappa\}|$? I'm having trouble understanding precisely what is this set $\{f(\xi): \xi < \kappa\$. Would you mind explaining your answer a bit? Thanks. –  Acid2 Sep 7 '12 at 15:12

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.