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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