Universally embedding total order = $\eta_\xi$ set?

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

• 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.) Apr 18, 2022 at 18:17
• I'm referring to this: en.wikipedia.org/wiki/%CE%97_set Apr 18, 2022 at 20:44
• 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. Apr 18, 2022 at 21:49

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.
• 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? Apr 19, 2022 at 8:23
• 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 $S$. Apr 20, 2022 at 6:37
• 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? Apr 20, 2022 at 6:37
• @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 $hg\supseteq f$. Apr 20, 2022 at 18:11
• (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 $\le\vert\mathcal{X}\vert$). Apr 20, 2022 at 18:14