
## 3.7 Cofinality

A cofinal subset $S$ of a well-ordered set $T$ is a subset $S \subset T$ such that $\forall t \in T \exists s\in S (t \leq s)$. Note that a subset of a well-ordered set is a well-ordered set (with induced ordering). Given an ordinal $\alpha$, the cofinality $\text{cf}(\alpha )$ of $\alpha$ is the least ordinal $\beta$ which occurs as the order type of some cofinal subset of $\alpha$. The cofinality of an ordinal is always a cardinal (this is clear from the definition). Hence alternatively we can define the cofinality of $\alpha$ as the least cardinality of a cofinal subset of $\alpha$.

Lemma 3.7.1. Suppose that $T = \mathop{\mathrm{colim}}\nolimits _{\alpha < \beta } T_\alpha$ is a colimit of sets indexed by ordinals less than a given ordinal $\beta$. Suppose that $\varphi : S \to T$ is a map of sets. Then $\varphi$ lifts to a map into $T_\alpha$ for some $\alpha < \beta$ provided that $\beta$ is not a limit of ordinals indexed by $S$, in other words, if $\beta$ is an ordinal with $\text{cf}(\beta ) > |S|$.

Proof. For each element $s \in S$ pick a $\alpha _ s < \beta$ and an element $t_ s \in T_{\alpha _ s}$ which maps to $\varphi (s)$ in $T$. By assumption $\alpha = \sup _{s \in S} \alpha _ s$ is strictly smaller than $\beta$. Hence the map $\varphi _\alpha : S \to T_\alpha$ which assigns to $s$ the image of $t_ s$ in $T_\alpha$ is a solution. $\square$

The following is essentially Grothendieck's argument for the existence of ordinals with arbitrarily large cofinality which he used to prove the existence of enough injectives in certain abelian categories, see [Tohoku].

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 $\alpha$. $\square$

Comment #3400 by Robert Furber on

I need to comment on the bibliography entry for the Tōhoku, but couldn't do that, so I'll do it here.

The word "algèbre" is currently misspelt as "algébre" in the title. In LaTeX, this means there should be a ?96?e, with a backtick, rather than \'e, with an apostrophe.

Comment #3401 by Robert Furber on

For some reason my last comment came out wrong. The ?96? should have been a \ followed by a backtick, the thing you get by pressing the top left key on the keyboard.

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