Can an ordinal be limit of a smaller set of ordinals?

Question

The Wikpedia article Regular cardinal contains the following weird sentence:

An infinite ordinal $\alpha$ is regular if and only if it is a limit ordinal which is not the limit of a set of smaller ordinals which set has order type less than $\alpha$.

I'm confused by the double-which grammar and I'm not sure if I understand the statement at all. If an ordinal is the limit of a set (like $\omega$ is the limit of the set of finite ordinals), how can this set be smaller then the ordinal itself? I thought the ordinal is the set.

And is there even a difference between "is smaller" and "has order type less than"?

Answer 1

Look at the example in the article : $\omega + \omega$ is the limit of $\omega+n$ with $n$ finite ordinal.

So $\omega + \omega$ is the limit of the set $S = \{ \omega + n\,|\, n\in \omega\}$, and $S$ is indeed smaller than $\omega+\omega$ since $S\simeq \omega$ as an ordered set.

In general if $\alpha$ is the limit of a set $S$ of ordinals, then $S$ must be well-ordered, so $S\simeq \beta$ for some ordinal $\beta$. The definition you highlighted says that if $\alpha$ is regular, you can never take $S\subset \alpha$ and $\beta<\alpha$.

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Of course this is in some sense optimal because for any $\alpha$, you can always take both $S = \alpha$ (but in this case $S\not\subset \alpha$), and $S = \{\alpha\}$ (but then $\beta = 1$).

Answer 2

Let me start with your last question.

There is an important difference between is smaller than and has order type less than. Two well-orders of different types can have the same cardinality. For example, $\omega$ has smaller order type than $\omega+1$, but both of these sets are countably infinite, so neither is smaller than the other. It’s true that if $\langle A,\le\rangle$ and $\langle B,\preceq\rangle$ are well-orders, and $|A|<|B|$, then the order type of $\langle A,\le\rangle$ is necessarily less than that of $\langle B,\preceq\rangle$, but the reverse implication is false in general.

Now consider the cardinal $\omega_\omega$. It’s a cardinal because if the ordinal $\alpha$ is less than $\omega_\omega$, then $|\alpha|<|\omega_\omega|$, i.e., because it has greater cardinality than any smaller ordinal. Since $\omega=\omega_0<\omega_\omega$, $\omega_\omega$ is clearly an uncountable cardinal. (It is in fact the $\omega$-th uncountable cardinal.) But $\omega_\omega$ is the limit of the sequence $\langle\omega_n:n\in\omega\rangle$. This increasing sequence has order type $\omega$:

$$\begin{array}{ccc} 0&1&2&3&4&\ldots&n&\ldots\\ \omega_0&\omega_1&\omega_2&\omega_3&\omega_4&\ldots&\omega_n&\ldots \end{array}$$

Thus, $\omega_\omega$ is the limit of a sequence of smaller ordinals (in fact cardinals) that has order type $\omega$, an order type that is less than $\omega_\omega$. This is precisely what it means to say that $\omega_\omega$ is not regular (and hence is singular).

Now consider the cardinal $\omega_1$. The order types less than $\omega_1$ are precisely the order types of the ordinals less than $\omega_1$. Suppose that $\alpha$ is an ordinal less than $\omega_1$ and $\langle\beta_\xi:\xi<\alpha\rangle$ is an $\alpha$-sequence of ordinals smaller than $\omega_1$. If $\beta$ is the limit of this sequence, then

$$\beta=\sup_{\xi<\alpha}\beta_\xi=\bigcup_{\xi<\alpha}\beta_\xi\;;$$

$\alpha$ is countable, being less than $\omega_1$, as is each of the sets $\beta_\xi$, so $\beta$ is the union of countably many countable sets and must therefore be countable. In particular, $\beta<\omega_1$. Thus, $\omega_1$ is not the limit of a set $S$ of smaller ordinals such that the order type of $S$ is less then $\omega_1$. That’s exactly what it means to say that $\omega_1$ is regular.

Answer 3

First, the only regular ordinals are cardinals. Not every cardinal is regular. The first singular (non-regular) cardinal is $\aleph_{\omega}$: $$\begin{align} \aleph_{\omega} & = \sup_{\alpha < {\aleph_{\omega}}} \alpha \\ &= \bigcup_{\alpha < {\aleph_{\omega}}} \alpha \\ &= \bigcup_{n < \omega} \aleph_{n} \\ \end{align}$$ (The "limit" of a set of ordinals is their supremum, which happens to be the union of the set.) $\aleph_{\omega}$ the union of the set $S = \{\aleph_n\mid\ n < \omega\}$ which consists of cardinals smaller than $\aleph_{\omega}$ and which is of a smaller size: $\lvert S \rvert = \omega = \aleph_0$. $S$ is not all of $\aleph_{\omega}$, but it's an unbounded subset of $\aleph_{\omega}$, so its union equals $\aleph_{\omega}$.