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I am new to both here and real analysis. It would be quite silly question to most of you guys. But I spent all day, seriously, to figure out. But I couldn't. My question is from the proposition of Royden. [p.17, Royden, 4th]

Proposition. Every nonempty open set is the disjoint union of a countable collection of open intervals.

My question is when we pick one interval with one rational number, what are endpoints of the interval. In my thinking, both endpoints would be included in original open set. But Royden tells me that both endpoints are not included in that. What am I missing? What are the both endpoints?

Ted
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microeconomics
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    Suppose that the original open set is $U=(0,1)\cup(2,3)$: you can write this as the union of the two open intervals $(0,1)$ and $(2,3)$, both of which have rational endpoints that are not in $U$. – Brian M. Scott Aug 29 '13 at 04:26
  • Then if we have U=(0, 10), do we only have one sub interval itself? – microeconomics Aug 29 '13 at 04:29
  • Yes, in that case only one interval is needed. On the other hand, you can use infinitely many. Let $$\mathscr{U}={(p,q):p,q\in\Bbb Q\text{ and }0\le p<q\le 10};;$$ then $\mathscr{U}$ is a countable family of open intervals whose union is $(0,10)$. – Brian M. Scott Aug 29 '13 at 04:31
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    Thanks a lot! But Brian, then the set U you made is not disjoint, right? I need to find much easier book for studying. – microeconomics Aug 29 '13 at 04:34
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    You’re welcome! That’s right, it’s not disjoint. If you require disjoint open intervals, then $(0,10)$ is the only possible decomposition of $(0,10)$. In fact there is a theorem that every open set in $\Bbb R$ can be written uniquely as the union of a countable (which includes finite) family of pairwise disjoint open intervals; see here. – Brian M. Scott Aug 29 '13 at 04:35
  • Thanks you so much. Now I can sleep tonight. Tomorrow I will try to study it again. Have a good night! – microeconomics Aug 29 '13 at 04:38

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Let $U \subset \mathbb{R}$ be a non-empty open set. Let $R=\{r_n\}$ be the set of rationals contained in $U$. Note that $R$ is countable.

For each $r_n$, let $I_n$ be the largest interval containing $r_n$ that is a subset of $U$. That is, $I_n = (\inf \{x | (x,r_n] \subset U \},\sup \{x | [r_n,x) \subset U \} )$. A little work will show that $I_n \subset U$, each $I_n$ is open and non-empty. This gives $\cup_n I_n \subset U$.

A little more work will show that for any two intervals $I_m,I_n$, either $I_m = I_n$ or $I_m \cap I_n = \emptyset$.

If $x \in U$, then $(x-\delta,x+\delta) \subset U$ for some $\delta>0$, and $(x-\delta,x+\delta)$ contains some element of $R$, say $r_k$. Then we must have $x \in I_k$. This gives $U \subset \cup_n I_n$ and so $U = \cup_n I_n$.

Now choose the unique members of $I_n$, call them $I_{n_k}$. Then we see that $U = \cup_n I_n = \cup_k I_{n_k}$, where $I_{n_k}$ are a collection of disjoint, open intervals.

copper.hat
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