Existence of stationary subset of $\omega_2$ with certain properties

Question

How can I show the existence of a stationary subset $X\subset\omega_2$ with the properties

1. $x\in X$ implies $cof(x)=\omega$
2. For every $\alpha<\omega_2$ the set $\{x\in X\mid x<\alpha\}$ is not stationary in $\alpha$.

assuming, besides $ZFC$, either $V=L$ or the square-principle?

Answer 1

Assume that $\square_{\omega_1}$ holds and let $(C_\alpha \mid \alpha \in \operatorname{Lim}(\omega_2))$ be a witnessing sequence, i.e.

• $C_\alpha \subseteq \alpha$ is club,
• $C_\alpha$ is countable, whenever $\operatorname{cf}(\alpha) = \omega$,
• $C_\beta = C_\alpha \cap \beta$, whenever $\beta$ is a limit point of $C_\alpha$.

Note that each $C_\alpha$ has order type $\le \omega_1$. And for $\beta < \omega_1$ let $\xi^\alpha_\beta$ be the $\beta$-th element of $C_\alpha$ in its strictly monotone enumeration (if it exists, otherwise let $\xi^\alpha_\beta = 0$).

Observe the following

1. If $\beta < \omega_1$ is a limit ordinal and $\xi^\alpha_\beta \neq 0$, then $\operatorname{cf}(\xi^\alpha_\beta) = \omega$.
2. There is some limit ordinal $\beta < \omega_1$ s.t. $\{ \xi^\alpha_\beta \mid \alpha < \omega_2 \}$ is stationary.
3. This set answers your question, i.e. it is a stationary subset of $E^{\omega_2}_{\omega} = \{ \xi < \omega_2 \mid \operatorname{cf}(\xi) = \omega \}$ that doesn't reflect.