Lexicographic Orderings of Aronszajn Trees are Aronszajn Lines I am trying to prove that every lexicographic ordering of a Aronszajn tree is a Aronszajn Line.
If $T$ is a tree, a lexicographic ordering of $T$ is defined as follows: For each $\alpha<\text{ht}(T)$, let $\prec_\alpha$ be a total order on $\text{Lev}_\alpha(T)$. If $x \in T$, let $D_x=\{y \in T: y\leq x\}$. If $\alpha\leq\text{ht}(x)$, let $x(\alpha)$ be the $\alpha$th element of $D_x$. If $x, y$ are incomparable, let $E(x, y)$ be the first $\alpha$ such that $x(\alpha)\neq y(\alpha)$. The lexicographic ordering defined by the orders $\prec_\alpha$ is defined as: $x \prec y$ iff $x<y$ or $x$ and $y$ are incomparable and $x(E(x, y))\prec_{E(x, y)}y(E(x, y))$.
A Aronszajn line is a total order of cardinality $\omega_1$ that contains no copies of $\omega_1$, no reverse copies of $\omega_1$ and no uncountable subsets order-isomorphic to subsets of $\mathbb R$. I am trying to prove that every lexicographic ordering of a Aronszajn tree is Aronszajn Line. I have proved that it will have cardinality $\omega_1$ and that it will contain no copy of $\omega_1$ nor reverse copies of $\omega_1$. It remains to show that no uncountable subset of the lexicographic order will be order-isomorphic to a subset of $\mathbb R$. I have looked for references, the handbook of Set Theoretic Topology won't help me in this step and every reference I find tells me to go to the handbook. Can someone help me?
Edit: I have noticed that my proof that there are no decreasing $\omega_1$ sequences is wrong. Can someone also help me in this step?
 A: Let $S$ be an uncountable subset of $T$, and let $D$ be a countable subset of $S$; it suffices to prove that $D$ is not dense in $S$. Since $D$ is countable, there is an $\alpha<\omega_1$ such that $D\subseteq\bigcup_{\xi<\alpha}\operatorname{Lev}_\xi(T)$. Now use a cardinality argument to show that there are $x,y\in S$ such that $\operatorname{ht}(x),\operatorname{ht}\ge\alpha$, $x(\alpha)=y(\alpha)$, and $x\prec y$.
Added: The proof that I know for the non-embeddability of $\omega_1$ and $\omega_1^*$ works equally well for both. Assume that $\langle x_\xi:\xi<\omega_1\rangle$ is either strictly $\prec$-increasing or strictly $\prec$-decreasing. For $\alpha<\omega_1$ let
$$L_\alpha=\{x\in\operatorname{Lev}_\alpha(T):|\{\xi<\omega_1:x\prec x_\xi\}|=\omega_1\}\;.$$
Show first that $L_\alpha\ne\varnothing$ for each $\alpha<\omega_1$. Then show that in fact $|L_\alpha|=1$ for each $\alpha<\omega_1$. HINT: If not, let $\alpha$ be minimal such that $|L_\alpha|\ge 2$, pick distinct $x,y\in L_\alpha$, and get a contradiction by showing that $x\prec y$ and $y\prec x$. Finally, use this result to show that $T$ has an $\omega_1$-branch, which is of course impossible.
