A question about $\aleph_1$-dense sets and the basis problem for uncountable linear orderings

Question

I have a question which I have been unable to find a reference for and which I explain as follows:

Recall that a set of reals $X$ is $\kappa$-dense if between any two real numbers there are exactly $\kappa$ elements of $X$. In 1971 Baumgartner presented a paper (All $\aleph_1$-dense sets of reals can be isomorphic) in which he proved that the following statement is consistent with ZFC:

(1) Every two $\aleph_1$-dense sets of reals are order-isomorphic.

It is also known that the above statement is a consequence of PFA (Proper Forcing Axiom). One proof of this fact by Todorcevic can be found in his 1989 book Partitions Problems in Topology.

I will now explain how this relates to the basis problem for the uncountable linear orderings:

Recall that a class $\mathscr{B}$ of linear orderings is a basis for a class of linear orderings $\mathscr{L}$ if $\mathscr{B}\subseteq\mathscr{L}$ and for any linear ordering $L$ we have that $L\in\mathscr{L}\iff$ there is a $B\in\mathscr{B}$ such that $B\preceq L$.

($B\preceq L$ just means that there is an order preserving map from $B$ into $L$).

In the particular case of uncountable linear orderings, a recent result of J. Moore tells us that there is a five element basis for the uncountable linear orderings (A five element basis for the uncountable linear orderings, Annals of Mathematics 163, 2005).

On this paper, statement (1) is presented as a consequence of PFA on theorem 1.1. What I cannot understand is the following part:

In particular if X is a set of reals of cardinality $\aleph_1$, then X serves as a single element basis for the class of uncountable separable linear orders.

which is apparently a consequence of statement (1). In his 2007 book Walks on Ordinals and Their Characteristics, Todorcevic even attributes to Baumgartner that the above statement is a consequence of PFA.

This seemed like an easy task, however, I have been unable to prove it. What I have tried to do is take an arbitrary set of reals $X$ of cardinality $\aleph_1$ and "extend" it to a set $Y$ such that $Y$ is $\aleph_1$-dense and $X\preceq Y$, then for any uncountable separable linear ordering, I would somehow find an $\aleph_1$-dense set that embeds into it, then the result would follow from statement (1).

Answer 1

As you noticed, it is enough to show that every uncountable set of reals $X$ contains a copy of an $\omega_1$-dense set of reals. WLOG, assume size of $X$ is $\omega_1$. Consider the set $X'$ of all two sided condensation points of $X$. Show that $X \backslash X'$ is countable. Conclude that between any two points of $X'$, there are $\omega_1$ many points of $X'$. Now you can easily embed $X'$ onto an $\omega_1$-dense set of reals.

Answer 2

Edit: I misunderstood the question. I'll leave this here anyway.

Let $\mathcal{I}$ list the (countably-many) open intervals in $\mathbb{R}$ with rational endpoints. If $X$ has size $\aleph_1$, then there is some $I\in\mathcal{I}$ such that $X\cap I$ has size $\aleph_1$. Fix such an $I$. For each $J\in\mathcal{I}$, let $f_J : I\to J$ be an order-isomorphism. Put

$$ Y = X\cup \bigcup\{f_J''(X\cap I) \;|\; J\in\mathcal{I}\} $$

Then $Y\supseteq X$ and $Y$ is $\aleph_1$-dense.