Order Isomorphisms of $\aleph_2$

Question

Here's what I already know:

1. We say that an infinite well-ordered set $A$ is order isomorphic to the set of natural numbers $\aleph_0$ if every initial segment of $A$ is finite.
2. We say that an uncountable well-ordered set $B$ is order isomorphic to the set $\aleph_1$ if every intial segment of $B$ is countable.

Now I am looking to fill in the following blanks for $\aleph_2$:

We say that a ________ well-ordered set $C$ is order isomorphic to the set $\aleph_2$ if every intial segment of $C$ is _________.

The motivation here is that every cardinal is really a special sort of ordinal. I want to know what makes $\aleph_2$ different from its preceding cardinals, in a general sense.

The first blank will need to be a descriptor for all sets with cardinality $\geq \aleph_2$. The second blank will need to describe any sort of set with cardinality $<\aleph_2$. These terms may not necissarily exist, but a mathematical definition would work just as well.

Answer 1

I suspect you won't be too happy with this example, but there's really nothing clever to do here:

Every well-ordering of cardinality at least $\aleph_2$ (or if you prefer, cardinality $>\aleph_1$) is isomorphic to $\aleph_2$ iff all of its proper initial segments have cardinality $\le\aleph_1$.

In general, suppose $\kappa$ is a cardinal; then a well-ordering $L$ of cardinality $\ge\kappa$ is order-isomorphic to $\kappa$ iff every proper initial segment of $L$ has cardinality $<\kappa$.

Note that "cardinality $<\aleph_1$" means "countable" and "cardinality $\ge\aleph_0$" means "infinite," so the example "an uncountable well-ordering is isomorphic to $\aleph_1$ iff every proper initial segment is countable" is indeed a special case of the above. I think it's important to note that there are no specific term for properties such as "cardinality $\ge\aleph_2$" or "cardinality $\le\aleph_1$" analogous to "uncountable" and "countable."

In the case of a successor cardinal $\kappa=\lambda^+$, this amounts to: a well-ordering $L$ of cardinality $>\lambda$ is order-isomorphic to $\kappa$ iff every proper initial segment of $L$ has cardinality $\le\lambda$.

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The point is that cardinals are initial ordinals: an ordinal $\alpha$ is an initial ordinal iff there is no bijection between $\alpha$ and any $\beta<\alpha$. Cardinals, set-theoretically, are just initial ordinals; the use of the word "cardinal" is merely suggestive of the context.

If $\alpha$ is an initial ordinal, then by definition any proper initial segment of $\alpha$ is of cardinality $<\alpha$ - remember that a proper initial segment of any ordinal $\gamma$ is in fact an ordinal smaller than $\gamma$ (downwards-closed sets of ordinals are ordinals!).