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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 $B$.

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    $\begingroup$ 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? $\endgroup$ Commented Jun 17, 2020 at 14:11
  • $\begingroup$ This is essentially about completeness of reals, but the result looks so trivial as to demand the deep property of completeness. $\endgroup$
    – Paramanand Singh
    Commented Jun 19, 2020 at 7:06
  • $\begingroup$ The same result does not hold for intervals in $\mathbb {Q} $. $\endgroup$
    – Paramanand Singh
    Commented Jun 19, 2020 at 7:07

1 Answer 1

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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. $\ \square$

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  • $\begingroup$ 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? $\endgroup$
    – twosigma
    Commented Jun 18, 2020 at 20:41
  • $\begingroup$ 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.) $\endgroup$ Commented Jun 18, 2020 at 20:52
  • $\begingroup$ Thanks for the detailed answer and additional interesting propositions. For Prop 5, I have seen the proof before so that is ok. $\endgroup$
    – twosigma
    Commented Jun 18, 2020 at 21:46
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    $\begingroup$ 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. $\endgroup$ Commented Jun 18, 2020 at 22:32

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