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This question has probably been asked. However, I am not interested in just getting the answer to it. Rather, I am interested in collecting as many different proofs of it which are as diverse as possible. A professor told me that there are many. So, I invite everyone who has seen proofs of this fact to share them with the community. I think it is a result worth knowing how to prove in many different ways and having a post that combines as many of them as possible will, no doubt, be quite useful. After two days, I will place a bounty on this question to attract as many people as possible. Of course, any comments, corrections, suggestions, links to papers/notes etc. are more than welcome.

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First proof that comes in mind: if $O$ is an open set and $x \in O$ then there exists an interval $I$ such that $x \in I \subset O$. If there exists one such interval, then there exists one 'largest' interval which contains $x$ (the union of all such intervals). Denote by $\{I_\alpha\}$ the family of all such maximal intervals. First all intervals $I_\alpha$ are pairwise disjoint (otherwise they wouldn't be maximal) and every interval contains a rational number, and therefore there can only be a countable number of intervals in the family. –  Beni Bogosel Mar 2 '13 at 0:12
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Since this is a big-list question, I am converting it to CW. –  robjohn Mar 2 '13 at 9:01
    
Oh, OK! Thanks! –  Orest Xherija Mar 3 '13 at 1:04
    
I cannot understand how each $I_a$ is disjoint. If one interval contains x and another interval also contains x, aren't they intersecting? –  James May 29 at 1:41

7 Answers 7

Here’s one to get things started.

Let $U$ be a non-empty open subset of $\Bbb R$. For $x,y\in U$ define $x\sim y$ iff $\big[\min\{x,y\},\max\{x,y\}\big]\subseteq U$. It’s easily checked that $\sim$ is an equivalence relation on $U$ whose equivalence classes are pairwise disjoint open intervals in $\Bbb R$. (The term interval here includes unbounded intervals, i.e., rays.) Let $\mathscr{I}$ be the set of $\sim$-classes. Clearly $U=\bigcup\mathscr{I}$. For each $I\in\mathscr{I}$ choose a rational $q_I\in I$; the map $\mathscr{I}\to\Bbb Q:I\mapsto q_I$ is injective, so $\mathscr{I}$ is countable.

A variant of the same basic idea is to let $\mathscr{I}$ be the set of open intervals that are subsets of $U$. For $I,J\in\mathscr{I}$ define $I\sim J$ iff there are $I_0=I,I_1,\dots,I_n=J\in\mathscr{I}$ such that $I_k\cap I_{k+1}\ne\varnothing$ for $k=0,\dots,n-1$. Then $\sim$ is an equivalence relation on $\mathscr{I}$. For $I\in\mathscr{I}$ let $[I]$ be the $\sim$-class of $I$. Then $\left\{\bigcup[I]:I\in\mathscr{I}\right\}$ is a decomposition of $U$ into pairwise disjoint open intervals.

Both of these arguments generalize to any LOTS (= Linearly Ordered Topological Space), i.e., any linearly ordered set $\langle X,\le\rangle$ with the topology generated by the subbase of open rays $(\leftarrow,x)$ and $(x,\to)$: if $U$ is a non-empty open subset of $X$, then $U$ is the union of a family of pairwise disjoint open intervals. In general the family need not be countable, of course.

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I like this answer LOTS. :-) And +1. –  coffeemath Mar 2 '13 at 1:30
    
This answer is very clear. It depends on the axiom of choice though, are there any constructive variants of this argument? –  Bunder Mar 2 '13 at 8:28
    
@Bunder: No, it doesn’t depend on the axiom of choice. The only place where you might even think that it did is when I showed that there are only countably many intervals, but the rationals can be explicitly well-ordered, so no choice is necessary even there. –  Brian M. Scott Mar 2 '13 at 12:24

In a locally connected space $X$, all connected components of open sets are open. This is in fact equivalent to being locally connected.

Proof: (one direction) let $O$ be an open subset of a locally connected space $X$. Let $C$ be a component of $O$ (as a (sub)space in its own right). Let $x \in C$. Then let $U_x$ be a connected neighbourhood of $x$ in $X$ such that $U_x \subset O$, which can be done as $O$ is open and the connected neighbourhoods form a local base. Then $U_x,C \subset O$ are both connected and intersect (in $x$) so their union $U_x \cup C \subset O$ is a connected subset of $O$ containing $x$, so by maximality of components $U_x \cup C \subset C$. But then $U_x$ witnesses that $x$ is an interior point of $C$, and this shows all points of $C$ are interior points, hence $C$ is open (in either $X$ or $O$, that's equivalent).

Now $\mathbb{R}$ is locally connected (open intervals form a local base of connected sets) and so every open set if a disjoint union of its components, which are open connected subsets of $\mathbb{R}$, hence are open intervals (potentially of infinite "length", i.e. segments). That there are countably many of them at most, follows from the already given "rational in every interval" argument.

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"connected => interval" also requires a proof (usually using the intermediate step "path-connected"). –  Martin Brandenburg Mar 4 '13 at 19:29
    
Path connected is not needed. It would be circular, as one needs connectedness of intervals to see that path-connected implies connected... If a set is not order convex (so x < z < y, x,y in A but z not in A), then we have an immediate disconnection. That intervals are connected follows from completeness of the order, and is a standard fact for ordered spaces. –  Henno Brandsma Mar 4 '13 at 20:22

A variant of the usual proof with the equivalence relation, which trades in the ease of constructing the intervals with the ease of proving countability (not that either is hard...):

  1. Define the same equivalence relation, but only on $\mathbb Q \cap U$: $q_1 \sim q_2$ iff $(q_1, q_2) \subset U$ (or $(q_2, q_1) \subset U$, whichever makes sense).
  2. From each equivalency class $C$, produce the open interval $(\inf C, \sup C) \subset U$ (where $\inf C$ is defined to be $-\infty$ in case $C$ is not bounded from below, and $\sup C = \infty$ in case $C$ is not bounded from above).
  3. The amount of equivalence classes is clearly countable, since $\mathbb Q \cap U$ is countable.
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Let $U\subseteq\mathbb R$ open. Is enough to write $U$ as a disjoint union of open intervals.
For each $x\in U$ we define $\alpha_x=\inf\{\alpha\in\mathbb R:(\alpha,x+\epsilon)\subseteq U, \text{ for some }\epsilon>0\}$ and $\beta_x=\sup\{\beta\in\mathbb R:(\alpha_x,\beta)\subseteq U\}$.

Then $\displaystyle U=\bigcup_{x\in U}(\alpha_x,\beta_x)$ where $\{(\alpha_x,\beta_x):x\in U\}$ is a disjoint family of open intervals.

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These answers all seem to be variations on one another, but I've found each one so far to be at least a little cryptic. Here's my version/adaptation.

Let $U \subseteq R$ be open and let $x \in U$. Either $x$ is rational or irrational. If $x$ is rational, define \begin{align}I_x = \bigcup\limits_{\substack{I\text{ an open interval} \\ x~\in~I~\subseteq~U}} I,\end{align} which, as a union of non-disjoint open intervals (each $I$ contains $x$), is an open interval subset to $U$. If $x$ is irrational, by openness of $U$ there is $\varepsilon > 0$ such that $(x - \varepsilon, x + \varepsilon) \subseteq U$, and there exists rational $y \in (x - \varepsilon, x + \varepsilon) \subseteq I_y$ (by the definition of $I_y$). Hence $x \in I_y$. So any $x \in U$ is in $I_q$ for some $q \in U \cap \mathbb{Q}$, and so \begin{align}U \subseteq \bigcup\limits_{q~\in~U \cap~\mathbb{Q}} I_q.\end{align} But $I_q \subseteq U$ for each $q \in U \cap \mathbb{Q}$; thus \begin{align}U = \bigcup\limits_{q~\in~U \cap~\mathbb{Q}} I_q, \end{align} which is a countable union of open intervals.

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He wanted a countable union of disjoint open intervals. If $U$ is simply an open interval, then $U=I_q=I_{\tilde{q}}$ for all $q,\tilde{q} \in \mathcal{Q}\cap U$. –  José Siqueira Oct 15 '13 at 16:15
    
That it is a disjoint union follows from the definition of $I_q$: if $x \in I_q \cap I_p$ then $I_q \cup I_p \subseteq I_q$ and $I_p$; hence if $I_q \neq I_p$ then $I_q \cap I_p = \emptyset$. Strictly speaking one should throw away all repeated $I_q$s, and one can definitely do this without destroying the countability of the union. –  Stromael Oct 21 '13 at 11:14

Let $U$ be an open subset of $\mathbb{R}$. Let $P$ be the poset consisting of collections $\mathcal{A}$ of disjoint open intervals where we say $\mathcal{A} \le \mathcal{A}'$ if each of the sets in $\mathcal{A}$ is a subset of some open interval in $\mathcal{A}'$. Every chain $C$ in this poset has an upper bound, namely $$\mathcal{B} = \left\{ \bigcup\left\{J \in \bigcup\bigcup C : I \subseteq J \right\}: I \in \bigcup\bigcup C\right\}.$$ Therefore by Zorn's lemma the poset $P$ has a maximal element $\mathcal{M}$. We claim that the union of the intervals in $\mathcal{M}$ is all of $U$. Suppose toward a contradiction that there is a real $x \in U$ that is not contained in any of the intervals in $\mathcal{M}$. Because $U$ is open we can take an open interval $I$ with $x \in I \subseteq U$. Then the set $$\mathcal{M}' = \{J \in \mathcal{M} : J \cap I = \emptyset\} \cup \left\{I \cup \bigcup \{J \in \mathcal{M} : J \cap I \ne \emptyset\}\right\}$$ is a collection of disjoint open intervals and is above $\mathcal{M}$ in the poset $P$, contradicting the maximality of $\mathcal{M}$. It remains to observe that $\mathcal{M}$ is countable, which follows from the fact that its elements contain distinct rational numbers.

Note that the only way in which anything about order (or connectedness) is used is to see that $I \cup \bigcup \{J \in \mathcal{M} : J \cap I \ne \emptyset\}$ is an interval.

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And yes, before you ask, I know this proof is silly. –  Trevor Wilson Mar 8 '13 at 2:21
    
I think this proof conceptually connects many things, including well-ordering, order theory and so on. So, I really benefited. –  user64066 Oct 22 '13 at 20:19
    
@user64066 Thanks, I'm glad to hear it. –  Trevor Wilson Oct 22 '13 at 20:30

I hope that this is right as this is a lemma i've thought of and i plan to use in a project due in several days and it somewhat generalizes the question asked:

Suppose that $U$ is a set of intervals in $\mathbb{R}$ (closed, open, semi-closed, etc.). Then there exists a set of disjoint intervals $V$ in $\mathbb{R}$ s.t. $\bigcup_{I\in U}I=\biguplus_{I\in V}I$. If non of the intervals are degenerate or $U$ is countable, then this set can be taken to be countable. And if they were all open, then we could take the segments of $V$ to be open (And also, if $U$ is countable then we won't be needing the Axiom of Choice).

proof: Let us order the elements of $U$: $U=\langle I_\beta\,|\,\beta\leq\alpha\rangle$ where $\alpha$ is the first ordinal of cardinality $|U|$. (If $U$ is countable then this doesn't require AC and from here on will be standard induction with a simple construction in the end for $\omega$).

I'll build $V_\beta=\langle J^\gamma_\beta\,|\,\gamma\leq\beta\rangle$ - a sequence of segments for all $\beta\leq\alpha$ such that every two sets in $V_\beta$ are either disjoint or equal and such that $\displaystyle{\bigcup_{\gamma\leq\beta}I_\gamma=\biguplus_{\gamma\leq\beta}J^\gamma_\beta}$ and $\forall\beta$, $\langle J^\beta_\gamma\rangle_{\gamma\geq\beta}$ is a non-descending sequence of sets ,by means of transfinite induction.

For $V_0$ take, $V_0=\langle I_0\rangle$. Suppose that we have built the required $V_\gamma$, $\gamma<\beta$ for some $\beta\leq\alpha$, then we will build $V_\beta$ in the following way: $\forall\gamma<\beta$, denote $\widetilde{J}_\gamma$=$\bigcup_{\gamma\leq\delta<\beta}J_\delta^\gamma$-still segments (non-decreasing sequence). If $I_\beta$ is disjoint of all $\widetilde{J}_\gamma$, taking $V_\beta\!=\!\langle \widetilde{J}_\gamma\,|\,\gamma<\beta\rangle\cup\{(\beta,I_\beta)\}$ would give us a sequence $\langle V_\gamma\,|\,\gamma\leq\beta\rangle$ satisfying the required conditions of it (the only non trivial thing is that pairs of $\widetilde{J}_\gamma$ are either disjoint of each other or are equal, but that is also quite trivial since if the contrary would have occurred, then $\exists\gamma_1<\gamma_2<\beta$ s.t. $\widetilde{J}_{\gamma_1}\neq\widetilde{J}_{\gamma_2}$ and $\widetilde{J}_{\gamma_1}\cap\widetilde{J}_{\gamma_2}\neq\emptyset$, but then, $\exists \beta>\delta_1\geq\gamma_1, \beta>\delta_2\geq\gamma_2$ s.t. $J^{\gamma_1}_{\delta_1}\cap J^{\gamma_2}_{\delta_2}\neq\emptyset$, meaning that either $J^{\gamma_1}_{\delta_2}= J^{\gamma_2}_{\delta_2}$ or $J^{\gamma_1}_{\delta_1}= J^{\gamma_2}_{\delta_1}$ thus, $\forall\beta>\epsilon\geq\delta_1,\delta_2$, $J^{\gamma_1}_{\epsilon}= J^{\gamma_2}_{\epsilon}$ and since we are talking here about non-decreasing sequences, this will contradict $\widetilde{J}_{\gamma_1}\neq\widetilde{J}_{\gamma_2}$). And if $I_\beta$ isn't disjoint of all $\widetilde{J}_\gamma$, Then we can take $J_\beta^\gamma=\widetilde{J}_\gamma$ for all $\gamma<\beta$ that don't intersect with $I_\beta$ and $J_\beta^\gamma=\bigcup_{\delta<\beta\text{ s.t. }\widetilde{J}_\delta\cap I_\beta\neq\emptyset}{\widetilde{J}_\delta}\cup I_\beta$ - segment for all of the other $\gamma\leq\beta$. Then again from the same arguments, $\langle V_\gamma\,|\,\gamma\leq\beta\rangle$ would satisfy the required conditions.

Finally, we can take $V=\{J_\alpha^\beta\,|\,\beta\leq\alpha\}$ to get what we wanted in the first place. And obviously, if our segments were all non-degenerate to begin with, from the way we constructed our set, all of the segments in $V$ will be non-degenerate (and thus of positive measure), but they are disjoint and so there is only a countable number of them. And if the segments in $U$ were all open, then obviously, so will the segments in $V$.$\square$

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