Is the union of finitely many open sets in an omega-cover contained within

some member of the cover?

Let $\mathcal{U}$ be an open cover of $\mathbb{R}$ (Standard Topology)

such that $\mathbb{R} \not \in \mathcal{U}$ and for any finite set $A$

there is a $U \in \mathcal{U}$ such that $A \subseteq U$. We call such an

open cover an $\omega$-cover. Can we show that for any finite set $B

\subset \mathcal{U}$, there is a $V \in \mathcal{U}$ such that $\cup B

\subseteq V$?

Ultimately I'm working on showing the following. Let $\langle

\mathcal{U}_n: n \in \mathbb{N} \rangle$ be a sequence of $\omega$-covers.

Can we 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 open cover of

$\mathbb{R}$?

My approach here was to use each $\mathcal{U}_n$ to cover $[-n,n]$, thus

eventually covering all of $\mathbb{R}$. Since $[-n,n]$ is compact and

$\mathcal{U}_n$ is a cover, $\mathcal{U}_n$ has a finite subcover. But

that's as far as I can get unless what I conjectured above is true.

## No comments:

## Post a Comment