# An open interval is not a disjoint union of two or more open intervals?

It seems intuitively clear that an open interval (like $$(a,b), (a, \infty), (-\infty,a)$$ or $$\mathbb{R}$$) cannot be written as a disjoint union of two or more (nonempty) open intervals, but I'm not sure how to prove this rigorously.

Here's my attempt:

I assume the result that open intervals are connected. To prove the result, I show that the disjoint union of two or more open intervals is disconnected. Take any open interval of this union; call it $$A$$. Let $$B$$ denote the union of the other open intervals. So, our set is the disjoint union $$A \cup B$$. We wish to show that this is disconnected, i.e. that $$\overline{A} \cap B$$ and $$A \cap \overline{B}$$ are empty. We first show that $$\overline{A} \cap B = \emptyset$$. If $$A = \mathbb{R}$$ then $$A \cup B$$ would not have been disjoint, so $$A \neq \mathbb{R}$$ and hence $$A$$ is bounded in one direction, so one of its endpoints is a real number. Then $$\overline{A}$$ contains this real number. But $$B$$ cannot contain any of the real endpoints of $$A$$, because otherwise it will intersect with points of $$A$$ (since $$B$$ is open). So $$\overline{A} \cap B = \emptyset$$. Next we show that $$A \cap \overline{B} = \emptyset$$. If this were nonempty, then some element $$b$$ of $$\overline{B}$$ is in the open interval $$A$$. Since the closure of a set is the set of all its limit points, this means every neighborhood of $$b$$ contains elements of $$B$$. But this is impossible, since, for example, take an open interval centered at $$b$$ and contained in $$A$$; this is disjoint from $$B$$.

• you can make it simpler... Suppose for instance that $(a,b) = (a,c) \cup (d,b)$, with $(a,c) \cap (d,b) = \emptyset$ and, of course $a<c<d<b$. What hapens to $c$? does it belong to the union? – PierreCarre Jun 17 '20 at 14:11
• This is essentially about completeness of reals, but the result looks so trivial as to demand the deep property of completeness. – Paramanand Singh Jun 19 '20 at 7:06
• The same result does not hold for intervals in $\mathbb {Q}$. – Paramanand Singh Jun 19 '20 at 7:07

Let $$(a, b)$$ be an open interval, and $$(c, d)$$ an open interval properly contained in $$(a, b).$$ Then $$a \leqslant c < d \leqslant b,$$ and $$a < c$$ or $$d < b.$$ The set $$(a, b) \setminus (c, d) = (a, c] \cup [d, b)$$ is not open, because it contains $$c$$ or $$d$$ or both, but it does not contain a neighbourhood of either. Therefore $$(a, b)$$ is not the disjoint union of $$(c, d)$$ with the union of any non-empty collection of open intervals.

## An application

(Proposition 4 is the promised application, while Proposition 5 is a by-product of the argument.)

Proposition 1. An open interval is not the disjoint union of an open interval and a non-empty open set.

Proof. See above. $$\ \square$$

Proposition 2. The union of a non-empty collection of open intervals with a non-empty intersection is an open interval.

Proof. Let $$\mathscr{I}$$ be a non-empty collection of open intervals containing a given point $$c \in \mathbb{R},$$ and let $$J = \bigcup\mathscr{I}.$$ In $$\overline{\mathbb{R}} = \mathbb{R} \cup \{+\infty, -\infty\},$$ let $$a = \inf J$$ and $$b = \sup J.$$ Then $$a \notin J$$ and $$b \notin J.$$ If $$a < x < b,$$ then $$c \leqslant x < b$$ or $$a < x \leqslant c,$$ and in either case $$x \in I \subseteq J$$ for some $$I \in \mathscr{I}.$$ Therefore $$J = (a, b).$$ $$\ \square$$

Proposition 3. If $$x \in U \subseteq \mathbb{R},$$ and $$U$$ is open, then $$U = J \cup W,$$ where $$x \in J,$$ $$J$$ is an open interval, $$W$$ is an open set, and $$J \cap W = \varnothing.$$

Proof. Let $$J$$ be the union of all open intervals $$I$$ such that $$x \in I \subseteq U.$$ By Proposition 2, $$J$$ is an interval $$(a, b).$$ Clearly, $$a \notin U$$ and $$b \notin U,$$ therefore $$U = (a, b) \cup (U \cap (-\infty, a)) \cup (U \cap (b, +\infty)),$$ so we can take $$W = (U \cap (-\infty, a)) \cup (U \cap (b, +\infty)).$$ $$\ \square$$

Proposition 4. An open interval is not the disjoint union of two non-empty open sets.

Proof. Let $$I$$ be an open interval, and suppose that $$I = U \cup V,$$ where $$U$$ and $$V$$ are disjoint non-empty open sets. Take any $$x \in U.$$ By Proposition 3, $$U = J \cup W,$$ where $$x \in J,$$ $$J$$ is an open interval, $$W$$ is an open set, and $$J \cap W = \varnothing.$$ Therefore $$I = (J \cup W) \cup V = J \cup (W \cup V), \text{ and } J \cap (W \cup V) = (J \cap W) \cup (J \cap V) = \varnothing.$$ This contradicts Proposition 1; so the hypothesis that $$I = U \cup V$$ is false. $$\ \square$$

Proposition 5. Every open subset of $$\mathbb{R}$$ is the union of a countable collection of pairwise disjoint open intervals.

Proof. Let $$U$$ be an open subset of $$\mathbb{R},$$ and let $$\mathscr{J}$$ be the collection of all maximal open subintervals of $$U.$$ By Proposition 3, $$U = \bigcup\mathscr{J},$$ and any two members of $$\mathscr{J}$$ with a non-empty intersection are equal. Because each member of $$\mathscr{J}$$ contains a rational number, $$\mathscr{J}$$ is countable. $$\ \square$$

• For the proof of Prop 2, I'm not sure why we have "in either case $x\in I \subseteq J$ for some $I \in \mathscr{I}$"? Could you please elaborate? – twosigma Jun 18 '20 at 20:41
• It was a bit terse. If $c \leqslant x < b,$ then because $b = \sup J,$ there exists $y \in J$ such that $x < y,$ therefore there exists an open interval $I \in \mathscr{I}$ such that $y \in I.$ We then have $c \in I$ and $y \in I$ and $c \leqslant x < y,$ therefore $x \in I \subseteq J.$ The proof for $a < x \leqslant c$ is similar, with the inequality signs going the other way round. (By the way, I'm afraid the proof of Proposition 5 is even more terse, and it may be more easily understood with reference to the proof of Proposition 3 than its statement.) – Calum Gilhooley Jun 18 '20 at 20:52
• Thanks for the detailed answer and additional interesting propositions. For Prop 5, I have seen the proof before so that is ok. – twosigma Jun 18 '20 at 21:46
• Like you, I imagined at first that a quite elaborate proof would be needed to answer the general case of your question, even though @PierreCarre's comment takes care of the case of two intervals without fuss. It was only after a long comfortable soak in the bath last night that I realised there was a simple proof. By that time I had dreamt up most of these other arguments, because my initial more elaborate idea had been to prove Prop. 4 from first principles and deduce Prop. 1. Something like that, anyway. It was only in bed this morning, unable to sleep, that I got it all straight in my head. – Calum Gilhooley Jun 18 '20 at 22:32