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Every open set of real numbers is the union of a countable collection of disjoint open intervals.

I understand the proof given on Royden's real analysis book page 42, I'm having a hard time picturing what does the intervals look like on the real line. Can anyone give me some idea on that?

Suppose $O$ is open, for each $x\in O$, there exists $y>x$ such that the open interval $(x,y)\subset O$ and define $b=\sup \{y:(x,y)\subset O\}$. Define $a=\inf\{z: (z,x)\subset O\}$, and $I_x=(a,b)$ then the proof basically going to show that the collection of all such intervals forms a union of $O$ and the collection has countably many such intervals and they are pairwise disjoint.

Take the open set (0,1) and consider $x=1/2$ then $b=1$ and $a=0$ then we have $I_x=(0,1)$, it seems like no matter what $x$ I choose the collection seems to only have 1 open interval in it which is essentially the open set (0,1) itself, I guess it does satisfies the requirement in the proof, but something just seems off on my understanding.

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    $\begingroup$ If you start with an open interval $O$ then there is only one interval in the collection and there is nothing to be done in this case. $\endgroup$ Aug 8, 2022 at 0:00
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    $\begingroup$ What if we say $O=(0,1)\cup (3,5)$, then the collections seem to be only of those two intervals, perhaps I should try to look into some open sets that don't look like open intervals? $\endgroup$
    – Remu X
    Aug 8, 2022 at 0:03
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    $\begingroup$ Right, exactly...if you have a disjoint collection of open intervals, then guess what...the proof 'finds' exactly that collection. Your understanding seems good to me so far. It's not really a very deep result so you might be looking for something that's not really there. $\endgroup$
    – SBK
    Aug 8, 2022 at 0:47
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    $\begingroup$ By the way, "countable" here includes finite $\endgroup$ Aug 15, 2022 at 22:37

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