
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 $\alpha$. $\square$

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