Question about a linear order Let $L$ be the set of countable limit ordinals. For each $\alpha \in L$, let $\langle \alpha_n : n < \omega \rangle$ be a strictly increasing cofinal sequence in $\alpha$. Define a linear order on $L$ by $\alpha \prec \beta$ iff the least $n < \omega$ for which $\alpha_n \neq \beta_n$ satisfies $\alpha_n < \beta_n$ (So $\prec$ is just the lexic order). I am trying to show the following:


*

*$L$ is not a countable union of well ordered subsets under $\prec$.

*Every uncountable suborder of $L$ contains a copy of $\omega_1$.
I don't see how to show any of these. Any help/hints would be appreciated. I should add that this is not a homework question - A mean friend gave this problem to me.
 A: For problem 1, suppose for the sake of contradiction that L is a countable union of well-ordered sets under $\prec$. Hence there exists a well-ordered set under $\prec$ that is stationary. Say $A$. For each $\alpha\in A$, let $Succ_A(\alpha)$ be $\min_{\prec} (A\backslash (\alpha)_{\preceq}), (\alpha)_{\preceq}= \{\gamma\in A: \gamma \preceq \alpha\}$. Consider $A_n=\{\alpha\in A: n \text{ is the least such that }\alpha_n<(Succ_A (\alpha))_n\}$. There are only countably many of such sets and they form a partition of $A$, therefore we might find a $n$ such that $A_n$ is stationary. Now the function $f: \alpha\in A_n \mapsto \alpha_n$ is regressive, hence by Pressing Down Lemma there exists a $A'\subset A_n$ stationary and $\xi$ such that $\forall \beta\in A'$ $\beta_n=\xi$. 
Since for each $\beta\in A'$, $\beta\restriction n\in \xi^{n}$, which is countable, so we can find a stationary $A''\subset A'$ and $\eta_0, \cdots, \eta_{n-1}$ such that for all $\gamma\in A'', \forall 0\leq j\leq n-1 \ \gamma_j=\eta_j, \gamma_n=\xi$. Now we have a problem, pick any $\alpha\prec\beta\in A''$, $Succ_A(\alpha)_n>\xi$ and $Succ_A(\alpha)\restriction n =\alpha\restriction n= \beta\restriction n$, but this means $\beta\prec Succ_A(\alpha),\beta \neq Succ_A(\alpha)$, which is a contradiction. 
