TITLE: How to understand the regular cardinal?
QUESTION [6 upvotes]: How to understand the regular cardinal? Could someone give me some  examples?

REPLY [25 votes]: 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:

Cofinality and its Consequences