I am reading [this](http://dx.doi.org/10.1016/0166-8641%2882%2990065-7) article in which two properties of open covers are described:



> 
> **$\gamma$-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)
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> 
> **$\gamma'$-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)
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> $\underline{Lim}A\_n=\{ x \in X : \exists n\_0 \in \omega \space \forall n \geq n\_0 \space x \in A\_n \}$
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> 


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!



Properties of open covers
Two things:


* 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.)


This says that solving the problem for a sequence "refining" a given one suffices.
* 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.


