# Any open subset of $\Bbb R$ is a countable union of disjoint open intervals

Let $$U$$ be an open set in $$\mathbb R$$. Then $$U$$ is a countable union of disjoint intervals.

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.

• 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. Mar 2, 2013 at 0:12
• Since this is a big-list question, I am converting it to CW.
– robjohn
Mar 2, 2013 at 9:01
• I cannot understand how each $I_a$ is disjoint. If one interval contains x and another interval also contains x, aren't they intersecting? May 29, 2014 at 1:41
• Yes but their union would then be another interval that contains an $I_a$ and therefore must be equal to it by maximality. Mar 15, 2015 at 16:36
• It's an index variable on the family $\{I_\alpha\}$. Mar 27, 2015 at 14:19

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 $$U=\bigcup_{I \in \mathscr{I}} I$$. 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 order-convex sets. (A set $$C\subseteq X$$ is order-convex if whenever $$u,v\in C$$ and $$u, then $$x\in C$$. Intervals and rays are always order-convex, and the converse is true if the order is complete.) It does take a little work to verify that these sets are open.

Alternatively, $$U$$ is the union of a family of pairwise disjoint intervals, where the intervals are allowed to have endpoints in the completion of the linear order.

In general the family need not be countable, of course.

• I like this answer LOTS. :-) And +1. Mar 2, 2013 at 1:30
• This answer is very clear. It depends on the axiom of choice though, are there any constructive variants of this argument? Mar 2, 2013 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. Mar 2, 2013 at 12:24
• Is it correct to write each equivalence class as $I_x = (\inf\{y \in U: y \sim x\}, \sup\{y \in U: y \sim x\})$? Nov 4, 2015 at 16:09
• @IntegrateThis. Let $S$ be a $\sim$-equivalence class. For any $x\in S$ there exists $r>0$ such that $(-r+x,r+x)\subset U$ so $\forall y\in (-r+x,r+x)\;(y\sim x).$ So $(-r+x,r+x)\subset \{y\in U:y\sim x\}=S.$ Sep 24, 2018 at 20:38

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 $$U \subseteq \mathbb{R}$$ 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.

• 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 $q,\tilde{q} \in \mathcal{Q}\cap U$. Oct 15, 2013 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. Oct 21, 2013 at 11:14
• no matter what proof i find, i keep having the same question. yours is the one i see more often, and i know it must be a stupid question, but: how is any maximal interval of our open set NOT the entire set itself? it seems that whatever point we take, the maximal interval containing it while still being in our set must be the set itself. i would appreciate it so, so much if anyone could help me with this, though it must be stupid to ask
– anne
Jun 15, 2022 at 18:35
• Hi @anne. Open sets in $\mathbb{R}$ don't need to be connected. Consider the union of two disjoint open intervals $U = (0,1)\cup(2,3)$, which is open (see, e.g., en.wikipedia.org/wiki/Open_set#Properties). There is no single open interval $I \subseteq U$ that covers the entire set $U$. So if I pick $q = 0.5 \in (0,1)$, then the maximal interval $I_q \ni q$ will simply be $(0,1)$. Hope that answers your question. Jun 21, 2022 at 9:20
• @Stromael that was it! it makes perfect sense now as they needn't be connected, i didn't pay attention as i read "set" and automatically thought "interval". thank you!
– anne
Jul 4, 2022 at 11:25

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.

• "connected => interval" also requires a proof (usually using the intermediate step "path-connected"). Mar 4, 2013 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. Mar 4, 2013 at 20:22

Let $$U\subseteq\mathbb R$$ open. It 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.

The intervals appearing in the union are disjoint in the sense that every time $$x,y\in U$$ with $$x, then either $$(\alpha_x,\beta_x)=(\alpha_y,\beta_y)$$ holds, or $$(\alpha_x,\beta_x)\cap(\alpha_y,\beta_y)$$ is empty. To see this, suppose $$(\alpha_x,\beta_x)\cap(\alpha_y,\beta_y)$$ has an element. We claim that $$[x,y]\subseteq U$$. (For if $$x with $$t\not\in U$$, then $$\beta_x\leq t\leq \alpha_y$$.)

But if $$[x,y]\subseteq U$$, then both $$\alpha_x and $$\alpha_y, and both $$y<\beta_x$$ and $$y<\beta_y$$. Hence $$\alpha_x$$ and $$\alpha_y$$ can be expressed as $$\alpha_x=\inf\{\alpha\leq x:(\alpha,x+\epsilon)\subseteq U, \text{ for some }\epsilon>0\},$$ $$\alpha_y=\inf\{\overline\alpha\leq x:(\overline\alpha,y+\overline\epsilon)\subseteq U, \text{ for some }\overline\epsilon>0\},$$ and these are the same; so then also $$\beta_x$$ and $$\beta_y$$ are the same.

• Are you sure the union is countable here? Sep 11, 2014 at 16:55
• @user10444: It is countable yes! If you agree that $\{(\alpha_x,\beta_x):x\in U\}$ is a disjoint family of open intervals then you can see it by choosing $r_x\in \mathbb Q\cap (\alpha_x,\beta_x)$ for all $x\in U$. Then $\{r_x:x\in U\}$ is countable right? Note that the intervals $\{(\alpha_x,\beta_x):x\in U\}$ are not distinct!
– P..
Sep 11, 2014 at 19:03
• Recall, we can "choose" such an $r_{x}\in(\alpha_{x},\beta_{x})$ such that $r_{x}\in\mathbb{Q}$ by the Axiom of Choice and by the density of $\mathbb{Q}$ in $\mathbb{R}$. Aug 30, 2017 at 20:29
• @Procore AoC isn't actually needed since $\mathbb{Q}$ can be explicitly well ordered Oct 26, 2023 at 7:15

This proof is an extended version of the nice proof proposed by Stromael and it serves best for beginners who want to understand every detail(that one that for any established mathematician logically seems trivial) of the proof.

$$\textbf{Proof:}$$

Let $$U \subseteq \mathbb{R}$$ be open and let $$x \in U$$. Then Either $$x$$ is rational or $$x$$ is irrational.

Suppose $$x$$ is rational, then define

\begin{align} I_x = \bigcup\limits_{\substack{I\text{ an open interval} \\ x~\in~I~\subseteq~U}} I,\end{align}

Claim: $$I_x$$ is interval, $$I_x$$ is open and $$I_x \subseteq U$$

Definition: An interval is a subset $$I \subseteq \mathbb{R}$$ such that, for all $$a in $$\mathbb{R}$$, if $$a,b \in I$$ then $$c \in I$$.

Now, consider any $$a such that $$a,b \in I_x$$. We want to show that $$c \in I_x$$.

Denote $$I_a$$ to be an interval such that $$x \in I_a$$ and $$a \in I_a$$. In other words $$I_a$$ is one of the intervals from the union $$I_x$$ that contains $$a$$. In the same way, let $$I_b$$ be the interval such that $$x \in I_b$$ and $$b \in I_b$$.

1. $$c=x$$: If $$c=x$$ then by construction of $$I_x$$, $$c \in I_x$$

2. $$c: If $$c then we have that either $$a or $$a. Since $$x \in I$$ for every open interval $$I$$ of the union $$I_x$$ (by construction of $$I_x$$ ), we have that $$x \in I_a$$ and $$x \in I_b$$. Since $$x \in I_a$$ then because $$I_a$$ is an interval $$c \in I_a$$ and hence $$c \in I_x$$. And since $$x \in I_b$$ then because $$I_b$$ is an interval $$c \in I_b$$ and hence $$c \in I_x$$. Thus, we concluded that $$c \in I_x$$.

3. $$c > x$$: If $$c>x$$ then we have that either $$a or $$x. Since $$x \in I$$ for every open interval $$I$$ of the union $$I_x$$ (by construction of $$I_x$$ ), we have that $$x \in I_a$$ and $$x \in I_b$$. Since $$x \in I_b$$ then because $$I_b$$ is an interval $$c \in I_b$$ and hence $$c \in I_x$$. As for the second case, note that since $$x \in I_b$$ we have that $$a \in I_b$$. But then, because $$I_b$$ is an interval we have that $$c \in I_b$$ and hence $$c \in I_x$$. Hence we concluded that $$c \in I_x$$.

This Proves that $$I_x$$ is an interval.

$$I_x$$ is open because it is union of open sets.

$$I_x \subseteq U$$ by construction.

Suppose $$x$$ is irrational, then by openness of $$U$$ there is $$\varepsilon > 0$$ such that $$(x - \varepsilon, x + \varepsilon) \subseteq U$$, and by the property of real numbers that for any irrational number there exists a sequence of rational unmbers that converges to that irrational number, there exists rational $$y \in (x - \varepsilon, x + \varepsilon)$$. Then by construction $$(x - \varepsilon, x + \varepsilon) \subseteq I_y$$. 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.

Now let's show that intervals $$\{I_q \} ~\ q \in U \cap \mathbb{Q}$$ are disjoint. Suppose there is $$i, j, \in U \cap \mathbb{Q}$$ such that $$I_i \cap I_j \neq \emptyset$$ then $$I_i \subseteq I_q$$ and $$I_j \subseteq I_q$$ for some $$q \in U \cap \mathbb{Q}$$

Hence we constructed disjoint intervals $$\{I_q \} ~\ q \in U \cap \mathbb{Q}$$ that are enumerated by rational numbers in $$U$$ and whose union is $$U$$. Since any subset of rational numbers is countable, $$\{I_q \} ~\ q \in U \cap \mathbb{Q}$$ is countable as well. This finishes the proof.

• What is the guarantee of the statement 'Denote $I_a$ to be an interval such that $x \in I_a$ '?
– user464147
Dec 30, 2018 at 14:47
• Also to me the first part of this answer seemed obscure, I tried to clarify it here math.stackexchange.com/questions/3040319/… Jan 17, 2019 at 9:28
• Most probably I am very late for this comment but still I will clarify. By construction $I_x$ is such that $x \in$ every $I$ that forms $I_x$. Then I say that $a \in I_x$ this implies that $a$ must be a member of one of the $I$s that form $I_x$. Why? Because if this is not the case then $a$ is not in $I_x$. Then I just take and call $I_a$ one of the $I$s that contain that $a$. Since $x \in$ every I $x$ will be a member of that particular $I$ that we defined as $I_a$Let me know if this is wrong and why. Sep 5, 2020 at 7:15

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.

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.

• And yes, before you ask, I know this proof is silly. Mar 8, 2013 at 2:21
• I think this proof conceptually connects many things, including well-ordering, order theory and so on. So, I really benefited. Oct 22, 2013 at 20:19
• @user64066 Thanks, I'm glad to hear it. Oct 22, 2013 at 20:30

$\mathbb{R}$ with standard topology is second-countable space.

For a second-countable space with a (not necessarily countable) base, any open set can be written as a countable union of basic open set.

Given any base for a second countable space, is every open set the countable union of basic open sets?

Clearly, collection of open intervals is a base for the standard topology. Hence any open set in $\mathbb{R}$ can be written as countable union of open intervals.

If any two of exploited open intervals overlap, merge them. Then we have disjoint union of open intervals, which is still countable.

Let $G$ be a nonempty open set in $\mathbb{R}$. Write $a\sim b$ if the closed interval $[a, b]$ or $[b, a]$ if $b<a$, lies in $G$.This is an equivalence relation, in particular $a\sim a$ since $\{a\}$ is itself a closed interval. $G$ is therefore the union of disjoint equivalence classes.

Let $C(a)$ be the equivalence class containing $a$. Then $C(a)$ is clearly an interval. Also $C(a)$ is open, for if $k\in C(a)$, then $(k-\epsilon, k+\epsilon)\subseteq G$ for sfficiently small $\epsilon$.

But then $(k-\epsilon, k+\epsilon)\subseteq C(a)$; so $G$ is the union of disjoint intervals. These are at most countable in number by Lindel$\ddot{\rm o}$f's theorem. This completes the proof.

Reference:G. De Barra, Measure theory and Integration, Horwood Publishing.

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 $V$.$\square$

The balls with radii $\frac{1}{n}$ and center at a rational number form a basis for the euclidean topology. This family is countable as it is a countable union of countable sets. We have found a countable basis, so we are done.

• Ah, but the intervals are requested to be disjoint, and your intervals aren’t disjoint. Oct 8, 2016 at 3:11

Essentially nothing differs here from the two previous responses which rely principally on the fact that $\mathbb{R}$ is locally connected. However I present here a proof that hopefully will feel accessible to readers with a slightly lower level of topological literary at the cost of appearing cumbersome to an expert.

Consider the connected components of $U$. $U_x \subseteq U$ is defined to be the connected component of U containing $x$ if $U_x$ is the largest connected subset of $U$ which contains $x$. Clearly by definition $U_x=U_v$ if $v \in U_x$. Therefore if $U_a \cap U_b \neq \varnothing$ then $U_a=U_b$. We see that $\{U_x\}_{x\in U}$ is a disjoint collection. Also it should be clear that $\bigcup \limits_{x\in U} U_x = U$.

Now we show that $\forall x$ $U_x$ is open. Let $y\in U_x \subseteq U$. Since $U$ is open there exists $\epsilon>0$ such that $(y-\epsilon, y+\epsilon )\subseteq U$. Sets of real numbers are connected iff they are intervals, singletons or empty. $(y-\epsilon,y+\epsilon)$ an interval hence it is connected. Therefore since $U_y$ is the largest connected subset of $U$ containing $y$ we must have $(y-\epsilon,y+\epsilon)\subseteq U_y =U_x$. This shows that $U_x$ is open for all $x$.

$U_x$ open and connected implies that $U_x$ must be an open interval.

Also $\mathbb{Q}$ dense in $\mathbb{R}$, so $\forall x\in U$, $U_x\cap \mathbb{Q}\neq \varnothing$ and $U_x=U_q$ for some $q\in\mathbb{Q}$. So we can write $\{U_x\}_{x\in U}=\{U_q\}_{q\in S}$ for some $S\subseteq \mathbb{Q}$. $\mathbb{Q}$ is countable so $S$ is at most countable.

In conclusion, we have just shown that the union of the connected components of $U$ is a disjoint union of open intervals that equals $U$ and is at most countable.

The proof that every open set is a disjoint union of countably many open intervals relies on three facts:

• $\Bbb R$ is locally-connected
• $\Bbb R$ is ccc
• The open connected sets in $\Bbb R$ are open intervals

Let $U\subseteq \Bbb R$ be open. Then there is a collection of disjoint, open, connected sets $\{G_\alpha\}_{\alpha\in A}$ such that $U=\bigcup_{\alpha\in A} G_\alpha$. Since $\Bbb R$ is ccc, the collection $\{G_\alpha\}$ is at most countable. Since the open connected sets $\Bbb R$ are open intervals, $\{G_\alpha\}$ is a countable collection of disjoint, open intervals.

The first two facts allow us to see some generalizations. Namely any open set in a locally-connected, ccc space is a countable disjoint union of connected open sets. This applies to any Euclidean space. Although open connected subsets of Euclidean space are more complicated than open intervals, they are still relatively well-behaved.

• What does "R is ccc" mean? Feb 24, 2015 at 8:39
• It means "satisfies the countable-chain condition" which really concerns itself with anti-chains. It means that every collection of disjoint open sets is at most countable. Every separable space is ccc.
– user123641
Feb 25, 2015 at 0:56
• Ok thanks for clearing that up. Feb 25, 2015 at 5:18
• how to decompose (0,1} into a countable union of disjoint open intervals? Mar 27, 2015 at 15:31

The following is certainly not the quickest approach to a proof, but when this question was first posed to me in class, my first intuition was to use some elementary graph theory:

Let $U$ be an open set of $\mathbb{R}$. As we know, $\mathbb{R}$ has a countable basis $\mathcal{B}$ comprised of connected open sets and so we may write $U=\bigcup_{n\in I} U_n$, where for each $n$ we have $U_n\in\mathcal{B}$ and $I$ is some countable index set.

Let $G$ be the intersection graph of $\{U_n\}$. That is to say, the vertex set of $G$ is simply $\{U_n\}$ and there is an edge between $U_i$ and $U_j$ iff they have nonempty intersection. It's easy to convince yourself that:

• This graph must have countably many graphically-connected components (otherwise we'd have uncountably many vertices which is impossible).
• The intersection graph of $A\subseteq\{U_n\}$ is graphically-connected iff for any two $V,W\in A$ there is a sequence $V=U_{n_1},U_{n_2},\ldots,U_{n_k}=W$ such that $U_{n_i}\cap U_{n_{i+1}}\neq\varnothing$.
• The union $\bigcup A$ is a connected set of $\mathbb{R}$ whenever the intersection graph of $A$ is graphically-connected.

Thus, when we take the union of all the vertices within a graphically-connected component, for every component, we obtain countably-many connected open sets. The union of these sets is of course $U$ itself. Since the connected open sets of $\mathbb{R}$ are intervals (including rays), we're done.

Side Note: This would also work in $\mathbb{R}^n$ or in general, any topological space $X$ that has a countable basis comprised of connected sets. Well, so long as we replace countable union of disjoint open intervals with countable union of disjoint open connected sets.

The balls with radii $\frac{1}{n}$ and center at a rational number form a basis for the euclidean topology. This family is countable since $\mathbb N \times \mathbb Q \equiv \mathbb N$ and we have a countable basis $(B_\lambda)_{ \, \lambda \in \mathbb N \times \mathbb Q}$ of open intervals for $\mathbb R$.

Let $U$ be a nonempty open set in $\mathbb R$; we can express it as countable union of open balls from $(B_\lambda)$. Also, $\tag 1 \text{ }$ $\quad$ If two open intervals have a nonempty intersection, then their union is also an open interval.

So if $U$ is a finite union of the $B_\lambda$, it is an easy matter to combine the $B_\lambda$, if necessary, and writing $U$ as a disjoint union of a finite number of open intervals. So assume, WLOG, that

$\tag 2 U = \bigcup_{\, n \in \mathbb N \,} B_n$.

We define a relation on our (new) index set $\mathbb N$ with $m\sim n$ if $B_m \cap B_n \ne \emptyset$ or there is is a finite 'nonempty intersection $B\text{-}$chain' connecting $B_m$ with $B_n$. It is easy to see that this partitions $\mathbb N$ and that taking the corresponding unions of the $B_n$ over an index $\lambda \text{-}$ block gives a partition of $U$. Also, using (1), we can show that we have also expressed $U$ as a countable union of disjoint open intervals.

More of a question than answer. I am a chemist turned pharmacist, who wishes to have studied mathematics. I am trying to work through Rudin's Principles of Mathematical Analysis. I envy you all who are involved in math for a career.

Can someone give me feedback on my attempt at a proof in \$\mathbb R? Completely novice and not at all pretty, but is it sound?

segment := open interval in R.

Lemma: disjoint segments in R are separated (proof not shown).

It follows from the lemma that an open connected subset of R cannot be the union of disjoint segments.

Let E be an open subset of R. Since R is separable by Rudin prob 2.22, there exists a subset, D, of R that is countable and dense in R. Assume E is connected, which includes E=R, then E is the union of an at most countable collection of open segments, containing only E.

Suppose E is separated. Then E is the union of a collection of disjoint segments, including the possibility of segments unbounded above or below.

If the collection is finite, then it is at most countable.

Assume the collection of segments is infinite. Because D is dense in R and E is contained in R, every open subset of E contains a point of D. Then each of the infinitely many disjoint segments contains a unique point of D. Since D is countable, a one-to-one correspondence between a unique point of D and the segment is established. This implies that there are countably many disjoint segments in the collection.

Therefore E is the union of finitely many or countably many , hence at most countably many, disjoint segments.

• More of a question than answer. I suppose this would be better posted as a question than as an answer, then. Suppose E is separated. Then E is the union of a collection of disjoint segments. It's not obvious (to me, at least) where this follows from in the given context.
– dxiv
Oct 9, 2016 at 5:43
• I know. Just learning how things work on this site. Which I think is quite amazing.
– RJM
Oct 9, 2016 at 5:48
• @dxiv Since E separated (or not connected), would it have to at least be the union of 2 separated sets. Since it is open, they have to be open subsets, which implies open interval in R1. Thank you!
– RJM
Oct 9, 2016 at 5:53
• @dxiv i will not keep bugging you. Any tips on how to become mathematician-like on one's own in essentially matgematica isolation?
– RJM
Oct 9, 2016 at 6:00
• IMHO that latter implies open interval in R1 is a part that needs to be proved in this context. See for example Intervals are connected and the only connected sets in R. Anyway, as I said, yours appears to be better suited as a question than as an answer. P.S. Keep plugging away until it all starts making sense ;-)
– dxiv
Oct 9, 2016 at 6:00