Progress on determining which partial orders embed into the rationals

Question

The following result is relatively well-known: (for example in Math StackExchange answer #37161)

For every countable linear order $(R,\prec)$, there is an $X\subseteq\mathbb Q$ such that $(R,\prec)$ is isomorphic to $(X,<)$, i.e. $X$ ordered by the usual order on rationals.

Generalizing this idea, the paper "Embedding Trees in the Rationals" (Baumgartner, Malitz, Reinhardt, Proceedings of the National Academy of Sciences vol. 67, no. 4, 1970) considered a weaker notion of embedding into the rationals. They define:

[$(P,\leq)$] is tree-like iff for each $z\in P$, $P_q=\{x\in P:x\leq q\}$ is linearly ordered. $P$ is a tree iff each $P_q$ is well-ordered.

...

$P$ is embeddable in the rationals iff there is a function $f:P\to\mathbb Q$ preserving strict order, i.e., such that $x<y$ implies $f(x)<f(y)$. [Blackboard bold added in this quote.] Note that $f$ is not required to be one-one.

The main result of the paper is that Martin's axiom implies that every tree $(P,<)$ of cardinality $\leq\aleph_1$ and with no uncountable chains embeds into the rationals, and that since this consequent implies the Suslin hypothesis, it is independent of ZFC.

At the end of the paper, there are three open problems listed. The first two are even more general, as they deal with partial orders in general embedding into the rationals:

1. Is it consistent with ZFC to assume that for any partial order $P$, if $\textrm{card }P=\aleph_1$ and every uncountable subset of $P$ contains an uncountable anti-chain, then $P$ can be embedded into the rationals?
2. (Galvin, unpublished) Is it consistent with ZFC to assume that for any partial order $P$, if $\textrm{card }P=\aleph_2$ and if every $P'\subseteq P$ with $\textrm{card }P'\subseteq\aleph_1$ can be embedded in the rationals, then $P$ can be embedded in the rationals?

All seem to be variations on the question of what partial orders embed into $(\mathbb Q,<)$. Has there been progress on this topic in the last 50 years?

Answer 1

The following simple counterexample to Question 1 can be found on p. 473 of the Milner–Pouzet paper cited below.

Let $P=\omega_1\times\omega_1$ with the strict partial order $$(x,y)\lt(x',y')\iff x\lt x'\text{ and }y\gt y'.$$ By the Erdős–Dushnik–Miller theorem, every uncountable poset contains either an uncountable antichain or an infinite chain. Since $P$ contains no infinite chains, every uncountable subset of $P$ contains an uncountable antichain.

$P$ is not the union of countably many antichains (i.e., is not embeddable in $\mathbb Q$) because an antichain in $P$ can have uncountable intersection with at most one subset of the form $\{x\}\times\omega_1$.

Question 2 seems to be much deeper, and I don't know anything about it. The nearest thing I've found to a reference is in a survey paper of Stevo Todorcevic, Combinatorial dichotomies in set theory, Bull. Symbolic Logic 17 (2011), 1–72. In that survey Conjecture 3.4 is called "Galvin's Conjecture", and it is analogous to Question 2, but it asks about decompositions of posets into chains rather than antichains. On the other hand Conjecture 4.3 (Rado's conjecture) is just Question 2 for the special case of trees, and is apparently an open problem, or was at the time of that survey. But I've only skimmed it, and may have missed or misunderstood something.

Order embeddings and antichain decompositions. I thank @Holo for informing me in a comment that the following proposition is a special case of a theorem of Đuro Kurepa, Ensembles ordonnés et ramifiés, Thèse, Paris 1935; Math. Belgrade 4 (1935), 1–138. See also E. C. Milner and M. Pouzet, Antichain decompositions of a partially ordered set, in: Colloq. Math. Soc. János Bolyai 60, Sets, Graphs, and Numbers (Budapest, 1991), pp. 469–498.

Proposition. A poset $P$ is embeddable in $\mathbb Q$ if and only if $P$ is the union of countably many antichains.

Proof. For the nontrivial direction, suppose $P$ is the union of antichains $A_1,A_2,A_3,\dots,A_n,\dots$. Recursively construct order-preserving maps $$f_n:A_1\cup A_2\cup\cdots\cup A_n\to\mathbb Q$$ such that the range of $f_n$ is finite (in fact has at most $2^n-1$ elements) and $f_n\subseteq f_{n+1}$. Then $\bigcup_{n=1}^\infty f_n$ is an embedding of $P$ into $\mathbb Q$.