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Two ordered sets have the same order type if there exists an order isomorphism between them. Now the set $X$ of all order types of countable totally ordered sets has cardinality $2^{\aleph_0}$; see here. But I'd like to find the cardinality of a smaller set.

Let us define an equivalence relation $\sim$ on $X$ as follows. If $a$ is the order type of $A$ and $b$ is the order type of $B$, then we say that $a\sim b$ if there exists an order isomorphism between $A$ and a subset of $B$, and an order isomorphism between $B$ and a subset of $A$. My question is, what is the cardinality of the set of equivalence classes of elements of $X$ under this equivalence relation?

Is it still $2^{\aleph_0}$, or is it smaller than that?

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    $\begingroup$ Well, there are definitely still $\aleph_1$ different equivalence classes (the countable ordinals), so you can't prove in ZFC that there are less than $2^{\aleph_0}$. $\endgroup$ – Eric Wofsey Oct 8 '18 at 19:35
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    $\begingroup$ @EricWofsey Either an answer showing that there are $2^{\aleph_0}$ equivalence classes, or an answer showing that there are $\aleph_1$ equivalence classes, would satisfy me. $\endgroup$ – Keshav Srinivasan Oct 8 '18 at 19:43
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    $\begingroup$ @EricWofsey I guess the question is whether the number of equivalence classes is (in ZFC) provably equal to $\aleph_1$, or provably equal to $\mathfrak c$, or neither. $\endgroup$ – bof Oct 8 '18 at 20:05
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    $\begingroup$ Nice question. A small remark: The order of the rationals is universal in the sense that any countable linear order embeds into it. An order is scattered if and only if it contains no copy of $\mathbb Q$. Clearly, you can limit your attention to scattered orders. This answer (and the references given there) is now relevant: in particular, you cannot have infinite families of incomparable orders (under embeddability). $\endgroup$ – Andrés E. Caicedo Oct 8 '18 at 22:05
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    $\begingroup$ @AndrésE.Caicedo Silly question: under AD, can there be a well-quasi-order which surjects onto $\mathbb{R}$? If not, that plus Laver's theorem, the AD-version of Burgess' theorem, and the fact that $V$ and $L(\mathbb{R})$ agree about this question shows that the answer is $\omega_1$ assuming large cardinals ... :P (But seriously, I don't see how to cook up a "really long" wqo under AD.) $\endgroup$ – Noah Schweber Oct 9 '18 at 0:29
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The cardinality is $\aleph_1$.

As noted by Andrés Caicedo, the nonscattered countable linear orders form a single $\sim$-class, so it suffices to explain why there are only $\aleph_1$-many $\sim$-classes of scattered types.

Rich Laver proved that the embeddability quasiorder of (countable) nonscattered types is a well-quasiorder. Call this quasiorder $W$. Let $P=W/\sim$ be the quotient poset. The poset $P$ has no infinite antichains or infinite properly descending chains. The task is to determine the size of $P$.

Recursively partition $P$ into levels: $L_0$ is the antichain of minimal elements of $P$, and for each $\alpha>0$, define $L_{\alpha}$ to be the antichain of minimal elements of $P-\bigcup_{\beta<\alpha} L_{\beta}$. Let $\gamma$ be the least ordinal that does not index any level. Then $|P| = |\bigcup_{\alpha<\gamma} L_{\alpha}| = |\gamma|$, since $\gamma$ is infinite and the levels are finite and partition $P$.

The question asks for the number of $\sim$-classes, which is the cardinality of $P$. We have just shown that this is $|\gamma|$. By Eric Wofsey's comment, $|\gamma|\geq \aleph_1$. To obtain a contradiction, assume that $|\gamma|>\aleph_1$. Choose an ordinal $\delta$ so that $\aleph_1<\delta<\gamma$. Level $L_{\delta}$ is not empty, since $\delta<\gamma$, so pick a countable linear order $\Lambda_{\delta}$ so that $(\Lambda_{\delta}/\sim)\in L_{\delta}$. Every level $L_{\varepsilon}$, for $\varepsilon <\delta$, contains some $\Lambda_{\varepsilon}/\sim$, where $\Lambda_{\varepsilon}$ is a countable linear order embeddable in $\Lambda_{\delta}$. (If there exists some $\varepsilon < \delta$ whose level $L_{\varepsilon}$ does not contain the $\sim$-class of any poset embeddable in $\Lambda_{\delta}$, then the least-indexed level with this property would have to contain $\Lambda_{\delta}/\sim$, contradicting either $\varepsilon <\delta$ or the disjointness of the levels.)

But now we have a countable linear order $\Lambda_{\delta}$ which has $|\delta|$-many order types of pairwise $\sim$-inequivalent, countable linear orders that are embeddable into it, namely the $\Lambda_{\varepsilon}$'s. And remember that $\delta$ is uncountable. We have now obtained a contradiction to Fraisse's Second Conjecture (also proved by Laver), which states that a countable scattered linear order has only countably many $\sim$-types of suborders. \\\\\

You will find the statements of all four of Fraisse's Conjectures on page 178 of Rosenstein's book Linear Orderings, Academic Press, 1982. The proof (due to Laver) of Fraisse's Second Conjecture is Theorem 10.52 of this book.

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