3
$\begingroup$

I am reading this 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)

$\gamma'$-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!

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.