# Is the union of finitely many open sets in an omega-cover contained within some member of the cover?

Mathematics Asked by objectivesea on September 23, 2020

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. Let $$X$$ be any infinite $$T_1$$ space, and choose distinct $$x_0,x_1in X$$. Then $${Xsetminus{x}:xin X}$$ is an omega cover of $$X$$, but $$X=(Xsetminus{x_0})cup(Xsetminus{x_1})$$ is not a subset of any $$Xsetminus{x}$$.

Answered by StevenClontz on September 23, 2020

Open $omega$-covers of $mathbb{R}$ may not have the property you desire.

Consider the following open cover $mathcal{U}$ of $mathbb{R}$: each set in $mathcal{U}$ is a disjoint union of $m$ open intervals of length $frac1n$ for some $n geq m geq 1$.

This is clearly an $omega$-cover of $mathbb{R}$: given any finite set $A = { a_1 , ldots , a_m } subseteq mathbb{R}$ find $n geq m$ large enough so that $| a_i - a_j | > frac1n$ for $i neq j$, and then take $U = bigcup_{i=1}^m ( a_i - frac{1}{2n} , a_i + frac{1}{2n} )$. Then $U$ belongs to $mathcal{U}$, and contains each $a_i$.

Next note that each set in $mathcal{U}$ is the disjoint union of finitely many pairwise-disjoint intervals, the sum of the lengths of which is $leq 1$. It follows that no set in $mathcal{U}$ includes both $(0 , 1 )$ and $( 10 , 11)$, which both belong to $mathcal{U}$.

Answered by user642796 on September 23, 2020