Picturing the proof of every open set of real numbers is a countable union of disjoint intervals.

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.

• 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. Aug 8, 2022 at 0:00
• 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? Aug 8, 2022 at 0:03
• 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.
– SBK
Aug 8, 2022 at 0:47
• By the way, "countable" here includes finite Aug 15, 2022 at 22:37