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In the opening sentences of chapter 3.5 on Baire's Theorem, Abbott writes:

The structure of open sets is relatively straightforward. Every open set is either a finite or countable union of open intervals.

I have looked backed through chapter 3 and I cannot find where he has proved this. Is it because the statement is completely trivial? If not, then it doesn't seem like Abbott's style to just assert something like this... I'm wondering if anyone could give me some clarification.

J. W. Tanner
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dnlwng
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    https://math.stackexchange.com/questions/318299/any-open-subset-of-bbb-r-is-a-at-most-countable-union-of-disjoint-open-interv – angryavian Jun 08 '20 at 03:09
  • He probably had the student prove in in an exercise. – fleablood Jun 08 '20 at 03:11
  • The question you linked is similar, but is slightly harder, because Abbott doesn't say anything about the intervals being disjoint. I have looked through all the exercises and can't find where it is proved. I'm wondering if there's just a one liner or simple reason that I'm missing. – dnlwng Jun 08 '20 at 04:03
  • Well, intuitively. A set is open so around every point of the set there is on open interval contained in the set. As the rationals are dense in the reals every such interval contains and can be represented by a rational number it contains. So there can't be any singletons and there are only be at most countably many disjoint intervals making the set. – fleablood Jun 08 '20 at 04:49

1 Answers1

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If $A$ is an open set then every point, $\alpha$ is an interior point. So if for every $\alpha$ there is an open interval $I_\alpha$ so that $\alpha \in I_\alpha \subset A$.

So $\cup_{\alpha\in A}I_\alpha \subset A$. But also $A = \cup_{\alpha \in A} \subset \cup_{\alpha\in A}I_\alpha$ so $A =\cup_{\alpha\in A}I_\alpha \subset A$.

Now the rationals are dense so for each $I_\alpha$ there are rational numbers in it.

So for each rational $q\in A$ there are (probably many) $I_\alpha$ so that $q\in I_\alpha$ but we consider the collection of collections of Intervals $U_q= \cup_{I_\alpha\ni q} I_\alpha$ And $A=\cup_{\alpha\in A} I_{\alpha} = \cup_{q\in A\cap \mathbb Q} U_q$.

Claim: for each rational $q\in A$, $U_q$ is single interval.

Once shown, we are done.

If $x\in U_q, y\in U_q$ and $x< y$ then there is an $I_\alpha$ so that $x\in I_\alpha$ and as $I_\alpha$ is an interval all $w: \min(q,x)\le w \le \max(q,x)$ we will have $w\in I_\alpha$. There is an $I_\beta$ so that $y\in I_\beta$. So for all $w: \min(q,y) \le w\le \max(q,y)$ we wil have $w\in I_\beta$. So for any $w: x \le w \le y$ then $\min(x,q)\le x \le w \le y\le \max(x,q)$ so $w\in U_q$. So by definition $U_q$ is an Interval.

Oh, I guess I have to show $U_q$ is open.

Also have to acknowledge I am allowing of unbounded intervals. And as obviously there will be cases where $q\ne r$ but $U_q = U_r$ that $A= \cup_{a\in K}U_k$ for some $K \subset A\cap \mathbb Q$ and $K$ being "at most countable" is countable or finite.

...

Okay $U_q$ is open... Let $x \in U_q$ the $x \in I_\alpha$ for some $\alpha \in A$ so ... yeah, that's obvious. $I_\alpha$ is open si there is in interval $I_x$ so that $x \in I_x \subset I_\alpha\subset U_q$ so $U_q$ is an open interval.

fleablood
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