Detail about an answer of Any open subset of $\mathbb{R}$ is a countable union of disjoint open intervals

Question

I saw an answer to the question that any open subset of R is a countable union of disjoint open intervals. The link of the beautiful answer is this. I understand the statement in the answer but it remains one question: how to prove $I_{q}$ and $I_{p}$ are disjoint? The answerer says in comment that if $x \in I_{p} \cap I_{q}$, then $I_{p} \cup I_{q} \subseteq I_{p},I_{q}$ according to the definition of $I_{p}$, hence $I_{p} \cap I_{q} \ne \emptyset$ implies $I_{p} = I_{q}$. I have no idea why is it like that. Any help will be greatly appreciated.

Answer 1

Let $U \subseteq \mathbb{R}$ be open and let $x \in U$ rational. $I_x$ is defined as follows:

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

Let $p, q \in U$ rationals. If $I_p \cap I_q \neq \emptyset$, as they are open intervals, then $I_p \cup I_q$ is an open interval, and it is contained in $U$ (because $I_p$ and $I_q$ are subsets of $U$ by definition). Now, as $p, q \in I_p \cup I_q$, then (by the definition of $I_p$ and $I_q$) $I_p \cup I_q \subseteq I_p$ and $I_p \cup I_q \subseteq I_q$. Therefore, $I_p = I_p \cup I_q = I_q$.

Let $p, q \in U$ rationals. As $I_p \cap I_q \neq \emptyset$ implies that $I_p = I_q$, then we conclude that if $I_p \neq I_q$, then $I_p \cap I_q = \emptyset$.

Answer 2

The definition of $I_q \subseteq U$ is essentially that it is the largest open interval containing $q$ that is contained in $U$.

Suppose that for $p, q \in U$, we have that $x \in I_p \cap I_q$. Let $y \in I_p$; we show that $y \in I_q$ (the converse follows by symmetry and these together will show $I_p = I_q$). By definition, there is some open interval $B$ such that $\{y, p\} \subseteq B \subseteq U$. Now, because $x \in I_p \cap I_q$, there is an open interval $C$ such that $\{p, x\} \subseteq C \subseteq U$ and an open interval $D$ such that $\{x, q\} \subseteq D \subseteq U$. Observe now that the collection $B \cup C \cup D$ (1) contains everything in $\{y, p, x, q\}$, (2) is contained in $U$, and (3) is an open interval (as the union of "overlapping" open intervals is an open interval). Since there is an open interval contained in $U$ that contains $y$ and $q$ together, then we have $y \in I_q$, as desired.

Answer 3

$I_q$ and $I_p$ are not always disjoint. The important thing is that, if they are not disjoint, then they are the same interval. There is a skipped step in that answer where you need to choose one representative from each set of "duplicated" intervals in order for the union to be a disjoint union.

The reason they are the same interval if they are not disjoint is that, if they (or indeed any two open intervals) overlap, then their union is again an open interval (you can prove this with some simple algebra on the inequalities satisfied by their endpoints). Since this union, call it $I_{pq}$, contains both $p$ and $q$, it must be contained in both $I_p$ and $I_q$ by the definitions of those sets. So we have containment in both directions, implying $I_p = I_{pq} = I_q$.