Statement: every nonempy open subset $U\subset \mathbb{R}$ is an (at most countable) disjoint union of open intervals.

The proof came from the course notes (page 8) of a measure theory course at the link below



First, verify that if $I$ and $J$ are intervals and $I\cap J\neq \emptyset$, then $I\cup J$ is an interval.

Given $x\in U$, let

$\alpha_x=\sup\{a:[x,a)\subset U\}$,

$\beta_x=\inf\{b:(b,x]\subset U\}$,


Verify that, for $x,y\in U$ either $I_x=I_y$ or $I_x\cap I_y=\emptyset$.

Indeed $x\sim y$ if $I_x=I_y$ is an equivalence relation on $U$. Hence, $U=\bigcup_{x\in U} I_x$ expresses $U$ as a disjoint union of nonempty intervals, say $U=\bigcup_{p\in P}I_p$ where $P$ is an index set and the $I_p$ are nonempty intervals. For each $q\in \mathbb{Q}\cap U$ there exists a unique $p_q$ such that $q\in I_{p_q}$. On the other hand, for each $p\in P$ there is a $q\in \mathbb{Q}\cap U$ such that $q\in I_p$. Thus, the mapping from $\mathbb{Q}\cap U$ to $P$ defined by $q\rightarrow p_q$ is onto. It follows that $P$ is at most countable.


I want to check my understanding of the two verifications that the author asked the reader to do. Here are my attempts.

  1. Verify that if $I$ and $J$ are intervals and $I\cap J\neq \emptyset$, then $I\cup J$ is an interval.

Recall that, for a subset $I$ of the real numbers to be an interval it must satisfy that $$ (\forall a,b \in I)(a<c<b\implies c\in I).$$ If $I\cap J\neq \emptyset$, then there exists $t,a_I,b_I,a_J,b_J$ such that $$ a_I<t<b_I\text{ and }a_J<t<b_J $$ where $a_I,b_I,t$ are all in $I$ and $a_J,b_J,t$ are all in $J$. We will keep such $t$ for later use.
For all $a,b\in I\cup J$, take any $y$ satisfying $a<y<b$, then either $y<t$ or $y>t$. If $y<t$, then $y\in (a_I,t)$. So $y\in I$. If $y>t$, then $y\in (t,b_J)$. So $y\in J$. Therefore, $y\in I\cup J$ and $I\cup J$ is an interval.

  1. Verify that, for $x,y\in U$ either $I_x=I_y$ or $I_x\cap I_y=\emptyset$.

I will prove that $I_x\cap I_y\neq\emptyset$, then $I_x=I_y$.
If $I_x\cap I_y\neq\emptyset$, then $I_x\cup I_y$ is an interval by we have just shown. Then $$ I_x\cup I_y= \Big(\max(\beta_x,\beta_y),\ \min(\alpha_x,\alpha_y)\Big) $$ By construction, $I_x=I_x\cup I_y =I_y$.

I only know about the basic topology and author clearly tried to avoid the notion of connected set. It would be great if someone could improve my proofs and help me with the terminology. For example, the extended interval will mess up the use of max and min operators.

  • $\begingroup$ Your proof for 1) fails when $I$ or $J$ is a singleton: $[a, a] = \{a\}$. $\endgroup$ – 0XLR Nov 17 '20 at 9:02
  • $\begingroup$ Thanks. How would you word this into the proof? $\endgroup$ – chuck Nov 17 '20 at 9:07
  • $\begingroup$ Your proof for 1) is fine as long as $I$ and $J$ intersect at least two points (why?). If they intersect at only one point, it is pretty easy to see $I \cup J$ is a interval immediately (why?). $\endgroup$ – 0XLR Nov 17 '20 at 9:08
  • $\begingroup$ @0XLR Aren't intervals open, so $I, J$ can't be singletons $\endgroup$ – stackex33 Nov 17 '20 at 9:09
  • $\begingroup$ @stackex33 At least in the proof of 1) the OP did not specify open intervals. So I assume closed intervals, clopen intervals, etc are possible. $\endgroup$ – 0XLR Nov 17 '20 at 9:09

You can simplify the first part as follows (there's no need to fix $a_I, b_I$ etc). Fix a $t\in I\cap J, a, b\in I\cup J, a < b$ and let $y\in (a, b)$. If $a, b\in I$ or $a, b\in J$, then $y\in I$ or $y\in J$ respectively by definition, so assume wlog $a\in I, b\in J$. Now, either $y\leq t\implies y\in (a, t]\subset I$ or $y\geq t\implies y\in [t, b)\subset J$, therefore $y\in I\cup J$.

It is fine to use $\max, \min$ with $-\infty<a<\infty\forall a\in\mathbb{R}$. But it is not entirely clear (at least the way you've written it) how $I_x = I_x\cup I_y = I_y$. I prefer the following. Without loss of generality, assume $x<y$, so that $[x, y]\subset U$ because $I_x\cup I_y$ is an interval contained in $U$.

Now, given $t\in I_x$, either $[x, t)\subset U$ or $(t, x]\subset U$ and in both cases (actually, there are three cases : $t<x, x<t<y, y<t$), it is clear that $[y, t)$ or $(t, y]$ is contained in $U$, hence $t\in I_y$. Repeat the argument to show that $I_y\subseteq I_x$.

  • $\begingroup$ Thanks. I also prefer your constructive reasoning. I was trying to show that if $I_x\neq I_x\cup I_y$, then it contradicts the optimality of $I_x$. $\endgroup$ – chuck Nov 17 '20 at 9:32

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