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!
$X = [0,\omega_1]$ satisfies $\mathsf{S}_1 ( \Omega , \Omega )$. For this I will apply the following:
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
I'll also apply the following fact (easily proved):
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:
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:
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