What are two disjoint stationary subsets of ω1? I know if cf(μ)≥ ω2 then two disjoint stationary subsets of μ are {α less than μ : cf(α)=ω} and {α less than μ : cf(α)=ω1}. But I'm not sure what two disjoint stationary sets of ω1 are. Any help is appreciated.
 A: There is no concrete example. They can be constructed using the axiom of choice. The relevant theorem is that any stationary subset of $\omega_1$ is the union of $\omega_1$ many disjoint stationary subsets. A nice exposition can be found in Jech.
A: As mentioned above AC is required to find two disjoint stationary in $\omega_{1}$ but the more concrete (which is of course not concrete, but worth mentioned) example I have seen is the following (due to Banach):
Consider an injection $f:\omega_{1}\to\Bbb R$ and for each $q\in\Bbb Q$ the sets $A_{q}=\{\alpha<\omega_{1}; f(\alpha)<q\}$
and $B_{q}=\{\alpha<\omega_{1}; f(\alpha)>q\}$. It is immediate
that for every $q\in\Bbb Q$, $A_{q},B_{q}$ are disjoint, thus
our goal is to find some $q\in\Bbb Q$ such that
$A_{q},B_{q}$ are stationary.
This can be done as follows: consider the sets
$$I=\{q\colon A_{q} \
\text{contains a cub set}\}, \ \ \ J=\{q\colon B_{q} \
\text{contains a cub set}\}. $$
Observe then that $\sup J\leq \inf I$ and that the set
$$C:=\left(\cap_{q\in J}B_{q}\right)\bigcap
        \left(\cap_{p\in I}A_{p}\right)$$
contains a cub set (as by definition of $I$ and $J$ it
contains a countable intersection of cub sets) and
that each element of $C$ lies in the
interval $\left[\sup J,\inf I\right]$. Since now a cub set
is uncountable it follows that $\sup J<\inf I$ and so we
can find $q\in\Bbb Q$ such that $q\not\in I\cup J$. It then
can be 'easily' seen
that $A_{q},B_{q}$ are stationary, which ends the proof.
