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(1)In set theory, what is the purpose for defining the concept of cofinality?is it that important?

(2)The concept of cofinality finally leads to 2 types of infinite cardinal, for which the first one is regular cardinal and another one is singular cardinal defined this way:

An infinite cardinal $\aleph_{\alpha}$ is regular if $cf (\aleph_{\alpha})=\aleph_{\alpha}$ and singular if $cf (\aleph_{\alpha}) < \aleph_{\alpha}$ . I was wondering here...what is the meaning of this and its consequences(the fact that it has 2 types of cardinal)?

Your explanation is very much appreciated. thanks ahead.

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As an example, in the theory of model categories, one often uses the small object argument to obtain desired factorizations (this is used in, for instance, proving the existence of the Quillen model structure on topological spaces); then the point is that the domains of the trivial cofibrations and fibrations are "small" in that homming out of them commutes with sufficiently long transfinite compositions of suitable maps. Here "sufficiently long" means that the ordinal has cofinality greater than some cardinal. –  Akhil Mathew Mar 4 '11 at 13:17
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@Akhil: You totally lost me. haha. –  Seoral Mar 5 '11 at 3:16

3 Answers 3

up vote 10 down vote accepted

The way I look at cofinality is: $$cf(\kappa) = \min\{|A| \colon A\subseteq\kappa \wedge \forall \beta<\kappa\exists \alpha\in A (\beta\le\alpha)\}$$ That is, the smallest size you need in order to be unbounded.

Clearly $cf(\kappa)\le\kappa$ for every cardinal, and for regular cardinals this is really an equality - you need to have $\kappa$ many elements in order to be unbounded, you also have that every set of cardinality $\kappa$ is unbounded.

On the other hand, suppose $\kappa$ is singular, namely $cf(\kappa)<\kappa$, that means that you can take $<\kappa$ many steps and still be unbounded.

Example:
Consider $\kappa=\aleph_\omega$, that is the first cardinal number which is larger than $\aleph_n$ for every finite $n$. We have that $\langle\aleph_n\colon n\in\omega\rangle$ is a cofinal sequence, as every ordinal below $\kappa$ is smaller than some $\aleph_n$, but this is merely a countable set while $\kappa$ is very much uncountable.

Now suppose I want to prove some property about ordinals below $\aleph_\omega$ which has a somewhat inductive property, that is if it is true at one point it will be true below it. This process requires me only countably many steps and in fact it can be done with a simple induction over the natural numbers.

On the other hand, had I wanted to do the same on $\aleph_1$ (assuming the usual models of ZFC where it is regular) I would have to ensure the process to continue over $\aleph_1$ many elements.

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Wow, thanks for your answer..... By the way, where do you took your definition of cofinality from? perhaps from a book? –  Seoral Mar 5 '11 at 3:15
    
@Seoral: No problem. As for the definition... I really can't remember, but I think it was in the course I took in set theory. Perhaps it was slightly different, but equivalent nonetheless. –  Asaf Karagila Mar 5 '11 at 4:33

Some interesting results regarding the cofinality of a cardinal are related to the exponential function of cardinals. For example we can prove that $\kappa^{cf(\kappa)}>\kappa$ or we can prove that $cf(2^\kappa)>\kappa$. Actually this last one (along with Cantor's $\kappa<2^\kappa$) happens to be the only restrain to the values of $2^\kappa$ when $\kappa$ is a regular cardinal. For example we know that $2^{\aleph_0}$ cannot be any cardinal with cofinality $\aleph_0$, so it cannot be $\aleph_\omega$, $\aleph_{\aleph_\omega}$ etc. Also, it's easy to see that the cofinality of every ordinal is a regular cardinal.

It can be proved in ZF that a cardinal $\kappa$ is singular exactly when there exists a cardinal $\lambda<\kappa$ and a family of sets $\{S_\xi : \xi<\lambda\}$ such that $|S_\xi|<\kappa$ and $\bigcup_{\xi<\lambda}S_\xi=\kappa$. This says that a singular cardinal $\kappa$ is a union of less than $\kappa$ subsets of $\kappa$, each with less than $\kappa$ elements. The size of every infinite cardinal depends solely on regular cardinals: Every infinite cardinal can be written as a regular sum of regular cardinals.

Intuitively you can say that when we pass to a regular cardinal we perform a bigger leap in size. For example $\aleph_0$ is a regular cardinal. It cannot be written as a finite union of finite sets. In a sense moving from the finite to the infinite makes a huge difference. Regular cardinals have different properties than singular cardinals because of this.

Using the axiom of choice one can show that every successor cardinal is regular but it is provable that we cannot prove in ZFC neither the existence of uncountable regular limit cardinals nor the consistency of such existence (interestingly enough we cannot prove in ZFC minus infinity neither the existence of infinite sets nor the consistency of such existence).

Regular limit cardinals (that are uncountable) are called weakly inaccessible. Strengthening this a bit, let's assume that an uncountable regular cardinal $\kappa$ is a strong limit (that is for every $\lambda<\kappa$ we have $2^\lambda<\kappa$). These cardinals are called (strongly) inaccessible cardinals. They are important metamathematically because at the height of such a cardinal the cumulative hierarchy is a model of ZFC. That is to say, these cardinals are natural bounds to the methods we have for creating new sets through the axioms, and thus assumptions of the existence of such cardinals create a "bigger" universe of set theory. We can further strengthen these results by assuming that a cardinal has other properties (combinatorial for example) and yield much stronger forms of cardinals. This gave rise to the theory of large cardinals which provides a measurement for the consistency strength of set theoretic statements.

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Set theory is such a mystery...thanks. –  Seoral Mar 5 '11 at 3:18
    
A very minor correction, the limitations on $2^\kappa$ are for regular cardinals. For $\kappa>\operatorname{cf}(\kappa)>\omega$ the limitations are imposed by the behavior of the continuum function below $\kappa$. There are interesting results coming from the Galvin-Hajnal theory. –  Asaf Karagila Aug 13 '11 at 15:37
    
@Asaf: Yes, I meant to correct that but I kept forgetting to do so. Also, sorry for the delayed reply but I just got access to my computer after a long time. –  Apostolos Aug 30 '11 at 16:46
    
@Welcome back to the civilized world, kid ;-) I also made a minor correction to the last paragraph. –  Asaf Karagila Aug 30 '11 at 16:56

We could also view the definition of cofinality as: for each von Neumann cardinal $\alpha$ $$cf(\alpha):= \min \{ \beta \mid \text{there exists a cofinal}~~ f: \beta \to \alpha\}$$

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Note that equations are written within double dollars like $$...$$ and \mid that spaces well is preferred over the pipe |. –  user21436 Mar 1 '12 at 2:40

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