# Properties of open covers

$\gamma$-property: If $\mathcal U$ is an open $\omega$-cover of $X$, then there sequence $\{ G_n : G_n \in \mathcal U\} \subset \mathcal U$ such that $\underline{Lim} G_n = X$.(page 153)

$\gamma'$-property: If $\mathcal U_n$ is a sequence open $\omega$-covers of $X$, then there sequence $\{ G_n : G_n \in \mathcal U_n \}$ such that $\underline{Lim} G_n = X$.(page 155)

$\underline{Lim}A_n=\{ x \in X : \exists n_0 \in \omega \space \forall n \geq n_0 \space x \in A_n \}$

In page 156, it is proved that, $\gamma$-property implies $\gamma'$ property. The general idea of the proof is clear to me except of one remark. It is mentioned at the end of page 155 that, "As we can suppose that $\mathcal U_{n+1}$ is a refinement of $\mathcal U_n$ for every $n \in \omega$, it is enough to prove that there is an infinite subsequence $\langle n_k : k \in \omega \rangle$ and a sequence $G_k \in \mathcal U_{n_k}$ with $\underline{Lim}G_k = X$".

I don't see why we can assume that $\mathcal U_{n+1}$ is a refinement of $\mathcal U_n$ for every $n \in \omega$.

Any help?

Thank you!

• Suppose that $\{ \mathcal{U}_n \}_{n \in \omega}$ and $\{ \mathcal{V}_n \}_{n \in \omega}$ are sequences of open $\omega$-covers and each $\mathcal{V}_n$ is a refinement of $\mathcal{U}_n$. If there is a sequence $\{ G_n \}_{n \in \omega}$ such that $G_n \in \mathcal{V}_n$ and $\underline{\mathrm{Lim}}_n G_n = X$, then we can find a sequence $\{ H_n \}_{n \in \omega}$ such that $H_n \in \mathcal{U}_n$ and $\underline{\mathrm{Lim}}_n H_n = X$. (Just choose $H_n \in \mathcal{U}_n$ including $G_n$ as a subset.)
• If $\mathcal{U}$ and $\mathcal{V}$ are two open $\omega$-covers, then $\{ G \cap H : G \in \mathcal{U} , H \in \mathcal{V}, G \cap H \neq \varnothing \}$ is also an open $\omega$-cover and is a refinement of both $\mathcal{U}$ and $\mathcal{V}$.
This implies that we can "refine" a sequence of open $\omega$-covers so that each successive cover in the new sequence refines those which come before it.