Are the closed ordinals (apart from $0$ and $1$) precisely the regular cardinals?

Question

Given a partially ordered set $P$, a collection of partially ordered sets, call it $\mathcal{Q}$, and an arbitrary function $f : P \rightarrow \mathcal{Q},$ we can form a new partially ordered set $\Sigma f$ as follows.

1. The elements of $\Sigma f$ are the pairs $(p,q) \in P \times \bigcup \mathcal{Q}$ such that $f(p) \ni q$.

2. Define $(p,q) \leq (p',q')$ iff either

   • $p < p'$ in the sense of $P$, or
   • $p=p',$ and letting $Q = f(p)=f(p'),$ we have that $q \leq q'$ in the sense of $Q$.

Note in particular that if $P$ is totally ordered and the elements of $\mathcal{Q}$ are totally ordered, then for all functions $f : P \rightarrow \mathcal{Q}$ we have that $\Sigma f$ is totally ordered. Similarly if 'totally ordered' is replaced by 'well-ordered.'

Question 1. Is there a standard name/notation for this construction? Where can I learn more?

Now call a collection $\mathcal{Q}$ of posets closed iff for all $P \in \mathcal{Q}$ and all $f : P \rightarrow \mathcal{Q}$ we have that $\Sigma f$ is order-isomorphic to an element of $\mathcal{Q}.$ Note also that an ordinal is not only a poset, but also collection of posets. Thus any given ordinal may or may not be closed. And since cardinals can be viewed as particular kinds of ordinals, any given cardinal may or may not be closed. In fact, it is easy to see that every infinite cardinal is closed. (Thank you, Asaf, for your correction.) Furthermore, every regular cardinal is closed.

Question 2. Are the closed ordinals (apart from $0$ and $1$) precisely the regular cardinals?

Answer 1

Your claim is wrong. It isn't true that cardinals are closed, it is true that regular cardinals are closed. In fact that gives an exact characterization of the closed ordinals.

First it is trivial that finite ordinals larger than $1$ are not closed. Now suppose that $\alpha+1$ is a successor ordinal, then $2\in\alpha+1$ and $\alpha\in\alpha+1$ then the map $f(0)=\alpha, f(1)=2$ witnesses that $\alpha+1$ is not closed.

We can generalize this to the case where $\delta$ is not a regular cardinal:

If $\delta$ is a limit ordinal and $\lambda=\operatorname{cf}(\delta)<\delta$ then we have:

1. $\lambda+1\in\delta$, we have $\{\delta_i\mid i<\lambda\}\subseteq\delta$ which is a cofinal sequence.
2. The map $f(i)=\delta_i$ for $i<\lambda$ and $f(\lambda)=1$ gives us an order type of $\delta+1\notin\delta$.

As for the name, I'm not familiar enough with order theory to answer that, but it seems that this is a generalized sum of orders, over $P$, I don't know for certain though.