It is well-known that every open set in $\mathbb{R}$ is a disjoint countable union of open intervals; is there an explicit construction of an open cover of the whole real line consisting of collections of countable disjoint open intervals?
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5$\begingroup$ I think trivially, the whole real line, $(-\infty, \infty)$ works for what you are requesting. $\endgroup$– N. OwadMay 18 at 0:47
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$\begingroup$ @N. Owad. Yes, that is what I thought first; but I think in the proof of the proposition that every open set in $\mathbb{R}$ is a disjoint countable union of open intervals, intervals are assumed to be of finite length; thus I am more interested in the non trivial case. $\endgroup$– Haoqing YuMay 18 at 0:59
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1$\begingroup$ By the way, if you carefully read the proof of the theorem "every open set in $\mathbb{R}$ is a disjoint countable union of open intervals", you will find these disjoint intervals are unique, i.e. $(-\infty, +\infty)$ is also the unique decomposition of $\mathbb{R}$. $\endgroup$– AsiganMay 18 at 1:00
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$\begingroup$ @Asigan. I see, thank you and I think that $(-\infty, \infty)$ might be the only option. $\endgroup$– Haoqing YuMay 18 at 1:02
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2$\begingroup$ Yeah, the well known theorem isn't true if you only allow bounded open intervals. $\endgroup$– Thomas AndrewsMay 18 at 1:06
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Suppose $]a,b[$ is an interval of the family covering $R$. The point $a$ is in $R$ but not in $]a,b[$. If $a$ were in another interval $]c,d[$, a nbrd of $a$ enclosed in $]c,d[$ should contain points of $]a,b[$, which is impossible. The family does't cover $R$.
