# How to understand the regular cardinal? [closed]

How to understand the regular cardinal? Could someone give me some examples?

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## closed as off-topic by Andrés E. Caicedo, Grigory M, user2345215, Ali Caglayan, amWhyJan 21 '15 at 0:23

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We can be of greater assistance to you if you make your question more specific. – Austin Mohr Jan 11 '12 at 8:16
According to the examples on Wikipedia the "easiest" example of a regular cardinal are the finite ordinals. – Rudy the Reindeer Jan 11 '12 at 8:28
What do you know about cofinality? – tomcuchta Jan 11 '12 at 10:02
A littele. I just want to know others'opinion to the regular cardinality. – Paul Jan 11 '12 at 10:57

Note: this whole answer assumes the axiom of choice, without it everything changes and the rules apply in a whole other way.

To understand regularity one must understand what is cofinality. Cardinals are usually regarded as initial ordinals, i.e. ordinals that have no bijection with a smaller ordinal. $\newcommand{\cf}{\operatorname{cf}}$

Ordinals are ordered sets, and they are well ordered, which means the order behaves like a ladder. Much like an actual ladder, one can skip several levels in a single step. Cofinality of an ordinal is the question what is the minimal number of steps that one has to take in order to ensure "reaching the top"?

While reaching the top of an infinite ladder is impossible in most cases, we instead regard "reaching the top" as "eventually passing every level", i.e. being unbounded.

So cofinality asks a question about subsets of the ordinal. Subsets of an ordinal are well ordered by the restriction of the order, and thus they are isomorphic to an ordinal themselves. One can re-ask the ladder question in a new way:

What is the least order type of an unbounded set?

In an informal language, this is exactly the definition of cofinality. It is not hard to see that this least order type has to be an initial ordinal - a cardinal. We can then denote it by $\cf(\alpha)$.

For example, $\cf(\omega)=\omega$, since every finite subset of $\omega$ is bounded, and every infinite subset of $\omega$ has the order type of $\omega$.

In a similar way, $\cf(\omega_1)=\omega_1$ since a subset of $\omega_1$ is unbounded if and only if it is uncountable, and uncountable subsets of $\omega_1$ are always of the same order type.

In fact, by Zermelo's theorem that $\aleph_\alpha\cdot\aleph_\alpha=\aleph_\alpha$ we have that every successor cardinal is regular, this by the very same considerations as we did with $\omega$ and $\omega_1$.

The ordinal $\omega_1+\omega$ has cofinality of $\omega$ since we can find an unbounded set of order type $\omega$, namely $\{\omega_1+n\mid n\in\omega\}$. However $\omega_1+\omega$ is not an initial ordinal, so comes the question can we have cardinals not equal to their cofinality?

Well, the answer is indeed yes. If a cardinal is the limit of smaller cardinals then it can have a strictly smaller cofinality than itself. For example, $\aleph_\omega=\sup\{\aleph_n\mid n\in\omega\}$ and the corresponding initial ordinals give us the unbounded set in $\omega_\omega$, so we have $\cf(\aleph_\omega)=\omega$.

In fact, one can easily check that: $$\cf(\aleph_\alpha)=\begin{cases}\aleph_\alpha & \alpha=\beta+1\\ \cf(\alpha) & \text{otherwise}\end{cases}$$

We finally reach the point of the answer: $\kappa$ is a regular cardinal if and only if $\kappa=\cf(\kappa)$. If $\kappa$ is a successor cardinal, then it is always regular, if it is a limit cardinal it is unprovable whether or not $\kappa$ can be regular. Note too, that since $\cf(\alpha)$ is a cardinal it is impossible an ordinal which is not a cardinal to be regular.

Why is it unprovable? If $\kappa=\aleph_\kappa=\cf(\kappa)$ (i.e. a regular limit cardinal) then we can produce a model in which there is no such cardinal. However a claim provable from ZFC must hold in all the models of ZFC. (Currently there is no known refutation of the existence of these cardinals either)

Further reading material on this site:

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Hi. Can you explain why $$\cf(\aleph_\alpha)=\begin{cases}\aleph_\alpha & \alpha=\beta+1\\ \cf(\alpha) & \text{otherwise}\end{cases}$$ I can see why it is true in the successor case but not the limit case. I get that $\cf(\aleph_\alpha)\le\cf(\alpha)$ but why can't it be strictly smaller? – Elie Bergman Jul 4 at 15:54
Because of the fact $\kappa\cdot\kappa=\kappa$ for infinite cardinals. Repeat the proof that $\aleph_1$ is regular to see that every successor is regular. For limits it follows from the fact that a limit cardinal is a limit of cardinals. – Asaf Karagila Jul 4 at 15:57

NOTE 1: I am not sure, whether there is some difference what happens if we are working in ZF only. I will assume Axiom of Choice in my post.

Cardinal $\kappa$ is regular if it cannot be obtained as union of system of less than $\kappa$ set, each of them having cardinality $<\kappa$. A cardinal, which is not regular, is called singular.

NOTE 2: Let us define here these notions for infinite cardinals only. (I am not sure what is the convention about finite cardinals. According to Wikipedia: Finite cardinal numbers are typically not called regular or singular.)

I am using one the usual definition of cardinals as initial ordinals - von Neuman construction, see e.g. this and this Wikipedia article. This means, that cardinal $\kappa$ is itself a set of cardinality $\kappa$. Another important thing is that in this representation the class of all cardinals is well-ordered.

With this representation we can show that a cardinal $\kappa$ is singular if it can be obtained as $\kappa=\sup\{\alpha_\gamma; \gamma<\kappa\}$ where $\alpha_\gamma<\kappa$ for each $\gamma<\kappa$. I.e., singular cardinal is cardinal which can be obtained as supremum of cardinals (ordinals) smaller than $\kappa$.

This shows how the notion of regular and singular cardinals is related to cofinality: An infinite cardinal $\alpha$ is regular if and only if $\alpha=\operatorname{cf}(\alpha)$. If is singular if and only if $\operatorname{cf}(\alpha)<\alpha$.

Examples:

• $\aleph_0$ is regular, since you cannot get $\aleph_0$ is finite union of finite sets.
• $\aleph_\omega$ is singular, since $\aleph_\omega=\sup\{\aleph_n; n<\omega\}$.
• It can be shown that every successor cardinal $\aleph_{n+1}$ is regular.

COMMENT: I didn't write here anything what cannot be found in Wikipedia article on regular cardinals or any standard set theory textbook. Maybe someone will come up with answer, which somehow gives you better intuition about these notions. But in case you checked this at Wikipedia or in some set theory textbook, it might be useful, if you specify what was unclear to you.

In case you want to learn something about regular and singular ordinals too, Wikipedia might be a good start.

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Maybe it need some time to read; however it looks nice! – Paul Jan 11 '12 at 8:35
In my answer I had the relation between $cf(\varkappa)$ and $\varkappa$ being regular wrong. I've edited the post to correct this. – Martin Sleziak Jan 11 '12 at 8:50
I've found it. So such as $\omega_2+\omega_1$ is singular? – Paul Jan 11 '12 at 8:53
The notation $\omega_2+\omega_1$ suggests that you mean ordinals and ordinal addition. In we work with ordinals, regular ordinal can be defined using $\operatorname{cf}(\alpha)=\alpha$. And $\omega_2+\omega_1$ is not regular. \\ If you meant cardinal addition then $\aleph_2+\aleph_1=\aleph_2$, which is regular. \\ If you are interested in regular ordinals as well, you should mention this in your question. – Martin Sleziak Jan 11 '12 at 8:57
There is a huge difference when not working in ZFC. Also regular ordinals are cardinals. – Asaf Karagila Jan 11 '12 at 9:43