The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

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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