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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.

Rushabh Mehta
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objectivesea
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2 Answers2

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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 $\mathcal{U}$.

user642796
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No. Let $X$ be any infinite $T_1$ space, and choose distinct $x_0,x_1\in X$. Then $\{X\setminus\{x\}:x\in X\}$ is an omega cover of $X$, but $X=(X\setminus\{x_0\})\cup(X\setminus\{x_1\})$ is not a subset of any $X\setminus\{x\}$.

Steven Clontz
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