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In Alling's "Foundations of Analysis on Surreal Number Fields," he writes

If $X$ is an $\eta_\xi$-set and if $Y$ is a [totally] ordered set of power not exceeding $\aleph_\xi$, then there exists an order-preserving map $f$ of $Y$ into $X$.

In other words, every $\eta_\xi$ set is universally embedding for every totally ordered set of cardinality $\leq \aleph_\xi$.

Is the converse true? Do we have that for every totally ordered set that is universally embedding for cardinality up to $\aleph_\xi$, that it is also an $\eta_\xi$ set?

EDIT: an $\eta_\xi$ set is a totally ordered set which has the property that, for every pair of subsets $L$ and $R$ such that everything in $L$ < everything in $R$, if $L$ and $R$ are of cardinality less than $\aleph_\xi$, then for all $l \in L$ and $r \in R$, there exists $z$ such that $l < z < r$. See also: https://en.wikipedia.org/wiki/%CE%97_set

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    $\begingroup$ What is an $\eta_\xi$ set? (Note that any order into which an $\eta_\xi$ set embeds is also $\le\aleph_\xi$-universally embedding, so if $\eta_\xi$-ness isn't preserved by passing to a larger linear order we get an immediate answer.) $\endgroup$
    – Noah Schweber
    Apr 18, 2022 at 18:17
  • $\begingroup$ I'm referring to this: en.wikipedia.org/wiki/%CE%97_set $\endgroup$
    – Mike Battaglia
    Apr 18, 2022 at 20:44
  • $\begingroup$ Ah! I see what you're saying now. You can embed an $\eta_\psi$ set into some larger set that isn't $\eta_\psi$ by just adding a bunch of junk elements off to the end or something. Such a set will still let the other sets embed into it, so the answer is no. $\endgroup$
    – Mike Battaglia
    Apr 18, 2022 at 21:49

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Turning my comment into an answer:

If $L$ is $\le\aleph_\xi$-universally embedding, then so is every linear order into which $L$ embeds. So a negative answer to the question follows from showing that linear orders containing $\eta_\xi$-sets need not be $\eta_\xi$-sets themselves. This can be done in many ways; for example, given an $\eta_\xi$-set $A$, consider the linear order $2\cdot A$ gotten by replacing each point in $A$ by a pair of adjacent points.

Incidentally, we can consistently do even better: there can consistently be $\le\aleph_\xi$-universally embedding linear orders without any $\eta_\xi$-subsets. Specifically, fix some enumeration $(L_\theta)_{\theta<\kappa}$ of linear orders such that each linear order of size $\le\aleph_\xi$ appears as some $L_\theta$ (and $\kappa$ is an ordinal). The sum $$X=\sum_{\theta<\kappa}L_\theta$$ is then trivially $\le\aleph_\xi$-universally embedding (although really what this shows is that that's the wrong notion of universal embedding to consider). Now $X$ only has an $\eta_\xi$-subset if there is an $\eta_\xi$-set of cardinality $\le\aleph_\xi$, and in general there need not be.

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  • $\begingroup$ Thanks @NoahSchweber. That stronger notion of embedding seems very interesting, where any embedding of $A$ into $X$ can be extended to an embedding of $B$ into $X$ that preserves $A$'s embedding whenever $A$ embeds into $B$. Does that give us some nice characterization in terms of $\eta$ sets, e.g. this notion of universal embedding is equivalent to having an $\eta_\xi$ subset or something like that? $\endgroup$
    – Mike Battaglia
    Apr 19, 2022 at 8:23
  • $\begingroup$ Hm, I do think this stronger notion of universal embedding is related to being an $\eta_\psi$ set, but I am not exactly clear on what the stronger notion is. The idea is that for any linearly ordered set $S$ you want to embed it into your universally embedding set $X$ in such a way that extensions of your order, let's call them $T$ with $S \subset T$, also embed into $X$ preserving the embedding on $S$. $\endgroup$
    – Mike Battaglia
    Apr 20, 2022 at 6:37
  • $\begingroup$ But is the idea behind "strongly universal embedding" for $X$ that for every embedding of $S$ into $X$, the embedding can be extended to an embedding of $T$ preserving $S$? Or just that, for any $S \subset T$, there always exists at least one embedding of $S$ into $X$ that can be extended to an embedding of $T$, in addition to other embeddings that potentially don't? $\endgroup$
    – Mike Battaglia
    Apr 20, 2022 at 6:37
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    $\begingroup$ @MikeBattaglia The idea is indeed the former: that every embedding can always be extended. The precise definition I have in mind is the following: say that $\mathcal{X}$ (a single structure) strongly universally embeds $\mathbb{K}$ (a class of structures of the same type) iff every element of $\mathbb{K}$ embeds into $\mathcal{X}$ and for every $\mathcal{A},\mathcal{B}\in\mathbb{K}$ and every pair of embeddings $f:\mathcal{A}\rightarrow\mathcal{X}$ and $g:\mathcal{A}\rightarrow\mathcal{B}$ there is an embedding $h:\mathcal{B}\rightarrow\mathcal{X}$ such that $hg\supseteq f$. $\endgroup$
    – Noah Schweber
    Apr 20, 2022 at 18:11
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    $\begingroup$ (This is related to the amalgamation property in the setting of Fraisse limits and ages.) E.g. as you say, the linear order $\mathbb{Q}$ does not strongly universally embed the class of all countable linear orders, even though every countable linear order embeds into $\mathbb{Q}$ (indeed no infinite structure $\mathcal{X}$ ever strongly universally embeds the class of all structures of cardinality $\le\vert\mathcal{X}\vert$). $\endgroup$
    – Noah Schweber
    Apr 20, 2022 at 18:14

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