Any open subset of $\Bbb R$ is a countable union of disjoint open intervals. [Collecting Proofs]

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

Lemma: Let $\{I_\lambda\}_{\lambda\in L}$ be a family of open inervals, such that $p\in I_\lambda$, for each $\lambda\in L$. Then $I=\cup I_\lambda$ is an open interval.

Proof: Suppose that $I_\lambda=(a_\lambda,b_\lambda)$. Note that $a_\lambda<b_\mu$, $\forall\lambda,\mu\in L$, because $a_\lambda<p$ and $p<b_\mu$. Therefore, if $a=\inf\{a_\lambda:\ \lambda\in L\}$ and $b=\sup\{b_\lambda:\ \lambda\in L\}$, then $a<b$. We claim that $(a,b)=\cup_\lambda I_\lambda$. Indeed, the inclusion $\cup_\lambda I_\lambda\subset(a,b)$ is clear. On the other hand, if $a<x<b$, then there exist $\lambda,\mu\in L$ such that $a_\lambda<x<b_\mu$. If $x<b_\lambda$ then $x\in I_\lambda$ and we are finished. If $x\geq b_\lambda$, then $a_\mu<b_\lambda\leq x$, which implies that $a_\mu<x<b_\mu$, i.e. $x\in I_\mu$ and this finish the Lemma's proof.

Now, suppose that $A$ is a open set of $\mathbb{R}$. For each $x\in A$, let $I_x$ be the union of open intervals that contain $x$ and that are contained in $A$. By using the Lemma each $I_x$ is a open interval. We claim that for $x,y\in A$, or either $I_x=I_y$, or $I_x\cap I_y=\emptyset$. Indeed, if there is some $z\in I_x\cap I_y$, then $I=I_x\cup I_y$ is an open interval containing $x$m which implies that $I\subset I_x$ and so $I_y\subset I_x$. By a similar argument, we get that $I_x\subset I_y$ and hence $I_x=I_y$. From what we seen, we can conclude that $A$ can be written as a union of open intervals.

Now, let's prove that this union is countable: For each $I_x$, take a rational nmber $r(I_x)\in I_x$. The function $I_x\mapsto r(I_x)$ is one-to-one, because $I_x\neq I_y$, implies $I_x\cap I_y=\emptyset$ which implies $r(I_x)\neq r(I_y)$. Because $\mathbb{Q}$ is countabe the result follows.

To finish, let's prove the unicity: Suppose that $A=\cup (a_m,b_m)$, where $(a_m,b_m)\cap (a_n,b_n)=\emptyset$ for $n\neq m$. We claim that $a_m,b_m\notin A$. Indeed, if for example $a_m\in A$, then $a_m\in (a_,b_p)$ for some $p\neq m$. Then, if $b=\min\{b_m.b_p\}$ we would have $\emptyset\neq (a_m,b)\subset(a_m,b_m)\cap (a_p,b_p)$ which is an absurd. It follows that for each $m$ and each $x\in (a_m,b_m)$, $(a_m,b_m)$ is the maximal open interval that's contains $x$ and is contained in $A$. Therefore $(a_m,b_m)=I_x$

Answer 6

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.

Answer 7

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.