## 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. 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.

Comment #7021 by R on

"The cofinality of an ordinal is always a cardinal (this is clear from the definition)."

I don't think this is immediate. It includes the statement that any countable limit ordinal has cofinality $\omega$, which to my knowledge requires a construction (of some cofinal subset). Or am I missing something?

Comment #7024 by on

@#7021: OK, I will remove the "this is clear from the definition" part next time I go through all the comments. OTOH, just now while having my morning coffee I wrote out why this is true and the only thing I needed to use was that if there is a bijection $\alpha \to \beta$ between an ordinal $\alpha$ and a cardinal $\beta$, then $\beta = \alpha$. Also, why in your comment you say one needs to construct a cofinal subset? The whole set itself is certainly cofinal, no?

Anyway, personally, I think of the cofinality as the least cardinal of a cofinal subset. But I think the definition in terms of ordinals is the one usually given in books, etc, which is why it is there.

Comment #7028 by R on

Recall your definition of a cardinal in 000D. In other terminology, the claim would be that any regular ordinal (one that occurs as the cofinality of something) is initial (a cardinal).

Comment #7030 by on

We have $\text{cf}(\alpha)$ is the least ordinal $\beta$ such that there is an order preserving bijection $\beta \to S$ where $S \subset \alpha$ is a cofinal subset. Define $\text{cf}'(\alpha)$ as the least cardinal $\kappa$ such that there is a bijection $\kappa \to S$ with $S \subset \alpha$ order preserving. If $\kappa = \text{cf}'(\alpha)$ and $\kappa \to S \subset \alpha$ is the bijection with a cofinal subset, then we may assume that $\kappa \to S$ is order preserving by what I said in my previous comment. Hence we see that $\text{cf}(\alpha) \leq \kappa = \text{cf}'(\alpha)$. Conversely, if $\beta = \text{cf}(\alpha)$ and $\beta \to S$ is an order preserving bijection, then set $\kappa$ equal to the cardinality of $\beta$ and you get a bijection $\kappa \to \beta \to S$. Since $\kappa \leq \beta$ we conclude that $\text{cf}'(\alpha) \leq \kappa \leq \beta = \text{cf}(\alpha)$. OK?

Comment #7031 by R on

There is a bijection between $\omega$ and $\omega \cdot 2$ (or $\omega^2$, or $\omega^\omega$, ...), but they don't all agree as ordinals. So I'm not sure what you're trying to say...

To show that the cofinality of $\omega \cdot 2$ is $\omega$, you chop off the first copy of $\omega$. For $\omega^2$ you use a diagonal, and for $\omega^\omega$ you do something similar.

Comment #7032 by on

Yes, I just realized what I was saying is nonsense. Sorry!

Comment #7033 by on

OK, the proof in Jech is very short (Lemma 3.8 on page 32).

Comment #7034 by on

OK, R, thanks and fixed here. The set theory chapter doesn't have proofs of most results, so it is fine to leave it unproven for now.

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