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Hi I see many people asked this question,but none of them proved this by considering all possible cases which my professor did.

Below is the solution that I have.

Show that a union of open intervals can be written as a disjoint union of open intervals.

Solution. The whole problem can be reduced to writing the union of two intervals as a union of disjoint ones (in case we have several, or uncountably many, we do induction). We are given (a, b) and (c, d) and we want to write

(a, b) ∪ (c, d) as a union of disjoint intervals. There are three cases that can occur: • a < b ≤ c < d. Then the two intervals are already disjoint, so we are done. • a ≤ c < b ≤ d. Then (a, b) ∪ (c, d) = (a, d). • a ≤ c < d ≤ b. Then (a, b) ∪ (c, d) = (a, b).

I am not sure why there are only three cases.

Can we have (c,d) and (c,b) as well? so I think there must be 5 cases that we have to consider. Also, I do not understand how I can apply this to show that any union of open intervals can be written as union of disjoint open intervals.

Can anyone explain?

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  • $\begingroup$ I'm actually a little confused by the outlined argument - the case analysis isn't perfect, as you point out, and the reduction to the two-interval case is not really that easy (to my mind). I would strongly prefer the standard "connected component" argument. $\endgroup$ – Noah Schweber Oct 28 '15 at 5:05
  • $\begingroup$ For the two interval problem, without loss of generality we may take $a\le c$. $\endgroup$ – André Nicolas Oct 28 '15 at 5:05
  • $\begingroup$ As @AndréNicolas pointed out, without loss of generality you can assume, that $a=\min\{a,b,c,d\}$ and $d=\max\{a,b,c,d\}$. But your proof works only if you have finite number of open sets, it fails even in the case on countable number of sets. $\endgroup$ – iiivooo Oct 28 '15 at 5:08
  • $\begingroup$ Hi I am little bit confused between open intervals and open sets. If we show that every open set is the union of disjoint open intervals, does that prove my problem? I am not sure whether Union of open intervals is equivalent to Open set $\endgroup$ – jessie Oct 28 '15 at 5:15
  • $\begingroup$ An open set is exactly a union of open intervals. $\endgroup$ – Noah Schweber Oct 28 '15 at 5:31
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Let $U \subset \mathbb{R}$ be open.

Define an equivalence relation $x \sim y $ iff $[\min(x,y), \max(x,y)] \subset U$.

Let $[x] \subset U$ denote the equivalence class containing $x$ (somewhat standard, but slightly ambiguous notation in this context).

Show that if $x \in U$, then $[x]$ is an open interval.

Let $\cal U = \{ [x] \}_{x \in U}$. Then $\cal U$ is a collection of disjoint open intervals such that $U = \cup_{A\in {\cal U}} A$.

Aside: By considering $U \cap \mathbb{Q}$, it is not difficult to show that the collection $\cal U$ is at most countable.

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  • $\begingroup$ Note that this proof is not the one the professor is outlining - in my opinion, it's better. $\endgroup$ – Noah Schweber Oct 28 '15 at 5:06
  • $\begingroup$ Hi I am little bit confused between open intervals and open sets. If we show that every open set is the union of disjoint open intervals, does that prove my problem? I am not sure whether Union of open intervals is equivalent to Open set $\endgroup$ – jessie Oct 28 '15 at 5:13
  • $\begingroup$ The union of an arbitrary collection of open sets is open. An open interval must be of the form $(a,b)$. An open set can be fairly 'wild', for example, $\cup_n B(q_n, {1 \over 2^n})$, where $q_n$ is an enumeration of the rationals. $\endgroup$ – copper.hat Oct 28 '15 at 5:17
  • $\begingroup$ @Garsa Yes, any open set is the union of open intervals (because open intervals form a so-called base for the topology). $\endgroup$ – Henno Brandsma Oct 28 '15 at 6:44

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