Does $X=[0,\omega_1]$ satisfy $S_1(\Omega,\Omega)$? Definition: An $\omega$-cover of a topological space $X$, is an open cover $\mathcal U$, such that, for any finite set $C \subset X$, there exists an open set $U \in \mathcal U$, such that, $C \subset U$.
Let $X=[0,\omega_1]$, where $\omega_1$ is the first uncountable ordinal.
Let $\langle \mathcal{U}_n: n \in \mathbb{N} \rangle$ be a sequence of open $\omega$-covers of $X$. Can we always find a sequence $\langle F_n: n \in \mathbb{N} \rangle$ with each $F_n \in \mathcal{U}_n$ such that $\cup F_n$ is an $\omega$-cover of $X$?
Any ideas or directions?
Thank you!
 A: $X = [0,\omega_1]$ satisfies $\mathsf{S}_1 ( \Omega , \Omega )$.  For this I will apply the following:

Theoerem (M. Sakai, 1988). A completely regular space $X$ satisfies $\mathsf{S}_1 ( \Omega , \Omega )$ iff $X^n$ satisfies $\mathsf{S}_1 ( \mathcal{O} , \mathcal{O} )$ for all $n \geq 1$

where a topological space $Y$ satisfies $\mathsf{S}_1 ( \mathcal{O} , \mathcal{O} )$ iff for any sequenec $\langle \mathcal{U}_n \rangle_{n \in \omega}$ of open covers of $Y$ there is a sequence $\langle U_n \rangle_{n \in \omega}$ such that


*

*$U_n \in \mathcal{U}_n$ for all $n$; and

*$\bigcup_{n \in \omega} U_n = Y$.


I'll also apply the following fact (easily proved):

Fact. Fix a base $\mathcal{B}$ for a topological space $Y$.  Then $Y$ satisfies $\mathsf{S}_1 ( \mathcal{O} , \mathcal{O} )$ iff it satisfies the requisite property for all sequences of open covers by sets in $\mathcal{B}$.

So I will show that $[0,\omega_1]^n$ satisfies $\mathsf{S}_1 ( \mathcal{O} , \mathcal{O} )$ for all $n < \omega$.  In fact, I will prove something somewhat stronger:
Proposition. Fix $n \geq 1$, and let $\langle \mathcal{U}_n \rangle_{n \in \omega}$ be a sequence of open covers of $[0,\omega_1]^n$ by "standard" basic open sets.  Then there is an $N < \omega$ such that one can pick $U_0 \in \mathcal{U}_0$, $\ldots$, $U_{N-1} \in \mathcal{U}_{N-1}$ with $[0,\omega_1]^n = U_0 \cup \cdots \cup U_{N-1}$.
(By "standard" basic open subsets of $[0,\omega_1]^n$ I mean products of open intervals.)
Proof sketch. I'll handle only the cases $n=1$ and $n=2$ in any detail.


*

*Let $\langle \mathcal{U}_n \rangle_{n \in \omega}$ be a sequence of covers of $[0,\omega_1]$ by open intervals. Setting $\alpha_0 = \omega_1$ we proceed as follows:


*

*If $\alpha_n$ is defined, pick $U_n \in \mathcal{U}_n$ containing $\alpha_n$.

*If $U_0 \cup \cdots \cup U_n \neq [0,\omega_1]$ set $\alpha_{n+1}$ to be least such that $( \alpha , \omega_1] \subseteq U_0 \cup \cdots \cup U_n \neq [0,\omega_1]$ (otherwise leave $\alpha_n$ undefined).


It trivially follows that $\alpha_0 > \alpha_1 > \cdots$, and so after finitely many steps it cannot be defined. But this means that $U_0 \cup \cdots \cup U_{N-1} = [0,\omega_1]$ for some $N < \omega$.

*Let $\langle \mathcal{U}_n \rangle_{n \in \omega}$ be a sequence of covers of $[0,\omega_1]^2$ by open intervals. Setting $\alpha_0 = \omega_1$ we proceed as follows:


*

*Set $\alpha_0 = \omega_1$, $N_{-1} = 0$

*If $\alpha_k$ is defined, note that as $[0,\omega_1] \times \{ \alpha_k \}$ is homeomorphic to $[0,\omega_1]$, and projections of sets in each $\mathcal{U}_n$ are open, by the above there is an $N_k > N_{k-1}$ such that for $N_{k-1} \leq i < N_k$ one may pick $U_i \in \mathcal{U}_i$ such that $[0,\omega_1] \times \{ \alpha_k \} \subseteq U_{N_{k-1}} \cup \cdots \cup U_{N_{k}-1}$.

*For each $N_{k-1} \leq i < N_k$ there is a least $\beta_i < \alpha_k$ such that $(\beta_i , \alpha_k] \subseteq \mathrm{proj}_2 (U_i)$ (if $[0,\alpha_k] \subseteq \mathrm{proj}_2 (U_i)$, leave $\beta_i$ undefined).  Let $\alpha_{k+1}$ be the maximum of the defined $\beta_i$ (if no $\beta_i$ is defined, leave $\alpha_{k+1}$ undefined).


As $\alpha_0 > \alpha_1 > \cdots$ it follows that after finitely many steps $\alpha_{k+1}$ cannot be defined, and it is relatively easy to show that $[0,\omega_1]^2 \subseteq U_0 \cup \cdots U_{N_k-1}$.

*(The general inductive step follows the case $n=2$ rather closely.) $\quad\Box$.

Reference

Masami Sakai, Property $\text{C}^{\prime\prime}$ and function spaces, Proc. Amer. Math. Soc. vol.104 (1988), no.3, pp.917–919, MR0964873, link

