The Stacks project

Proposition 3.7.2. Let $\kappa $ be a cardinal. Then there exists an ordinal whose cofinality is bigger than $\kappa $.

Proof. If $\kappa $ is finite, then $\omega = \text{cf}(\omega )$ works. Let us thus assume that $\kappa $ is infinite. Consider the smallest ordinal $\alpha $ whose cardinality is strictly greater than $\kappa $. We claim that $\text{cf}(\alpha ) > \kappa $. Note that $\alpha $ is a limit ordinal, since if $\alpha = \beta + 1$, then $|\alpha | = |\beta |$ (because $\alpha $ and $\beta $ are infinite) and this contradicts the minimality of $\alpha $. (Of course $\alpha $ is also a cardinal, but we do not need this.) To get a contradiction suppose $S \subset \alpha $ is a cofinal subset with $|S| \leq \kappa $. For $\beta \in S$, i.e., $\beta < \alpha $, we have $|\beta | \leq \kappa $ by minimality of $\alpha $. As $\alpha $ is a limit ordinal and $S$ cofinal in $\alpha $ we obtain $\alpha = \bigcup _{\beta \in S} \beta $. Hence $|\alpha | \leq |S| \otimes \kappa \leq \kappa \otimes \kappa \leq \kappa $ which is a contradiction with our choice of $\alpha $. $\square$


Comments (0)

There are also:

  • 12 comment(s) on Section 3.7: Cofinality

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 05N3. Beware of the difference between the letter 'O' and the digit '0'.