# Set of various order types of a set

Starting from the cardinal $$|\Bbb N| = \aleph_0 = \beth_0$$, we can generate a larger cardinal in two ways:

1. Take the set of all subsets, generating the cardinal $$\beth_1$$
2. Take the set of all well-order-types (up to isomorphism), generating the cardinal $$\aleph_1$$

I am wondering if, rather than well-order-types, we can take other order-types (up to isomorphism) to generate different cardinals. Are any of these known?

1. The set of all total order-types of $$\aleph_0$$
2. The set of all partial order-types of $$\aleph_0$$

3. The set of all preorder-types of $$\aleph_0$$

where these are all taken up to isomorphism.

Are the cardinalities of any of these known? Do any lead to another type of cardinal similar to the $$\aleph$$ or $$\beth$$ cardinals? How much of AC is required to talk about these?

• Since total orders are special types of partial orders, 2 and 3 are the same. – Noah Schweber Feb 9 '19 at 22:36
• Oops, that's from an earlier edit. Removed – Mike Battaglia Feb 9 '19 at 22:37

There's an easy construction which lets us show that there are continuum-many countable ordertypes up to isomorphism: given any infinite sequence of natural numbers $$f$$, consider the order $$A_f=\mathbb{Z}+f(0)+\mathbb{Z}+f(1)+...=\sum_{i\in\mathbb{N}}[\mathbb{Z}+f(i)].$$ It's easy to check that $$A_f\cong A_g\iff f=g$$. This immediately implies that all the other numbers are also continuum.
And note that we can do this for arbitrary infinite cardinalities: given $$f:\kappa\rightarrow\{0,1\}$$, let $$A^\kappa_f=\sum_{\eta<\kappa}[\mathbb{Z}+f(\eta)].$$ Again we have $$A_f^\kappa\cong A_g^\kappa$$ iff $$f=g$$.
So for every infinite cardinal $$\kappa$$, there are $$2^\kappa$$-many isomorphism types of linear orders of size $$\kappa$$. (And AC was never used at any point in the above.)
• @MikeBattaglia Not a single number, but a finite linear ordering of length $f(i)$, between copies of Z. – Ned Feb 9 '19 at 23:36
• @MikeBattaglia There is a natural way to view the set of all linear orderings of $\mathbb{N}$ as a Polish space $\mathcal{L}$. In this light, Burgess' theorem says that any analytic equivalence relation on $\mathcal{L}$ has either countably many, $\aleph_1$-many, or continuum-many equivalence classes. What does this have to do with counting orderings of a given type? Well, given a class $C$ of linear orders (e.g. the well-orderings), let $\equiv_C$ be the equivalence relation on $\mathcal{L}$ given by pushing all the "non-$C$" orders together, and being isomorphism otherwise. (cont'd) – Noah Schweber Feb 10 '19 at 0:44
• That is, $A\equiv_CB$ iff (neither $A$ or $B$ is in $C$) or ($A\cong B$). So for example: taking $C$ to be the class of well-orderings, $\mathcal{L}/\equiv_C$ has one element for each countable ordinal, and then one more element for "not well-ordered." So we're just counting the $\equiv_C$-classes, essentially. The point is that Burgess tells us that we can never hope to get anything other than $\le\aleph_0$, $\aleph_1$, or $2^{\aleph_0}$ using any kind of ordering which yields an analytic equivalence relation. For example, $C$=well-orderings does give an analytic equivalence relation. – Noah Schweber Feb 10 '19 at 0:47
• (And Silver's theorem, which is easier, shows that coanalytic equivalence relations are even tamer: they get either countably many or continuum-many classes.) So tentatively, I'd say that you're never going to find a natural way to whip up (say) $\aleph_2$ this way (even after granting the clearly-necessary assumption of $\neg$CH). – Noah Schweber Feb 10 '19 at 0:48
• All this falls into the general descriptive-set-theoretic picture of cardinals in between $\aleph_0$ and continuum being "difficult to exhibit." Incidentally, things get weirder at higher cardinals, but that's a separate issue. – Noah Schweber Feb 10 '19 at 0:51