Open sets can be expressed as a countable, disjoint, sub-collection of intervals.

Question

I am studying Capinsky and Kopp's book on Measure theory. For some reason, I went back to Chapter $1.$ where they make the following claim:

Definition $1.1.$

A subset $O$ of the real line $\mathbb{R}$ is open if it is a union of open intervals, i.e. for intervals $(I_{\alpha})_{\alpha \in > \Lambda},$ where $\Lambda$ is some index set (countable or not) $$O = \bigcup_{\alpha \in \Lambda} I_\alpha.$$

$\cdots \cdots \cdots$

If $\Lambda$ is an index set and $I_\alpha$ is an open interval for each $\alpha \in \Lambda$, then there exists a countable collection $(I_{\alpha_k} )_{k \geq 1}$ of these intervals whose union equals $\cup_{\alpha \in \Lambda}I_\alpha.$ What is more, the sequence of intervals can be chosen to be pairwise disjoint.

The last line marked in bold is where I have a problem. The fact that it can be expressed as a countable, disjoint union, of some intervals is known and has been dealt with on SE in multiple posts like in here or here just to name a couple. But the authors here make a different claim. Given a certain index set $\Lambda$, you can find a countable subset of this index set, say $\Gamma$ (so that $\Gamma \subseteq \Lambda$) such that $$\bigcup_{\beta \in \Gamma}I_\beta = \bigcup_{\alpha \in \Lambda}I_\alpha$$

and $I_{\beta_j} \cap I_{\beta_k} = \emptyset$, $\beta_j,\beta_k \in \Gamma,$ $i \neq j.$

Counter example:

Consider the set $O = O_1 \cup (-3,1) \cup (-1,3) \cup O_2.$ Here, $O_1$ and $O_2$ are open sets disjoint from $(-3,1) \cup (-1,3)$, and are such that $O_1 \cup O_2 = \bigcup_{\alpha \in \Lambda}I_\alpha$ for some uncountable index set $\Lambda$. No matter how you choose your sub-collection, if you have to cover the points $-2$ and $2$ which are in $O = O_1 \cup (-3,1) \cup (-1,3) \cup O_2,$ you h_ave to include $(-3,1)$ and $(-1,3)$ as $O_1 \cup O_2$ is disjoint from each of those sets. But $(-3,1)$ and $(-1,3)$ are not disjoint!

Where am I going wrong in this counter example?

Answer 1

The statement

If $\Lambda$ is an index set and $I_\alpha$ is an open interval for each $\alpha \in \Lambda$, then there exists a countable collection $(I_{\alpha_k} )_{k \geq 1}$ of these intervals whose union equals $\cup_{\alpha \in \Lambda}I_\alpha.$ What is more, the sequence of intervals can be chosen to be pairwise disjoint.

is plainly false. Your counterexample is correct. I shall come back to this later.

However, as the links in your question show, each open subset of $\mathbb R$ can be written uniquely as the countable union of disjoint open intervals. Note that "countable" includes "finite". As an example an open interval $(a,b)$ which is the disjoint union of exactly one open interval.

Thus, also the notation $(I_{\alpha_k} )_{k \geq 1}$ is a bit misleading because one is tempted to understand that there are infinitely many intervals. The same holds for "... the sequence of intervals can be chosen ...". We must be aware that "sequence" can mean a finite sequence $\alpha_1, \ldots, \alpha_n$.

Concerning counterexamples: Your counterexample is a bit abstract, it does not really specify an uncountable family of open intervals. Here is such a family:

$$\Lambda = (0,1), I_\alpha = (\alpha, \alpha+1) .$$ No two of these intervals are disjoint, thus there is no subset $\Gamma \subset \Lambda$ such that $\bigcup_{\beta \in \Gamma}I_\beta = \bigcup_{\alpha \in \Lambda}I_\alpha$ and $I_{\beta_j} \cap I_{\beta_k} = \emptyset$, $\beta_j,\beta_k \in \Gamma,$ $j \neq k$.

Perhaps the authors had in mind the following correct theorem:

For each family $(I_\alpha)_{\alpha \in \Lambda}$ of open intervals there exists a countable (possibly finite) subset $\Gamma \subset \Lambda$ such that $\bigcup_{\beta \in \Gamma}I_\beta = \bigcup_{\alpha \in \Lambda}I_\alpha$.

Here is a proof:

Let $C$ denote the set of all closed intervals $[a,b]$ which have rational endpoints $a, b$ and are contained in some $I_\alpha$. Since $\mathbb Q \times \mathbb Q$ is countable, also $C$ is countable (and, by the way, infinite). For each $c \in C$ choose $\alpha(c)$ such that $c \subset I_{\alpha(c)}$. Then $\Gamma = \{\alpha(c) \mid c \in C \}$ is a countable (possibly finite, though $C$ is infinite) subset of $\Lambda$. We claim that $$\bigcup_{\beta \in \Gamma}I_\beta = \bigcup_{\alpha \in \Lambda}I_\alpha .$$ The inclusion $\subset$ is trivial. So let $x \in \bigcup_{\alpha \in \Lambda}I_\alpha$. This means $x \in I_{\alpha_x}$ for some $\alpha_x$. But clearly there is $c \in C$ such that $x \in c \subset I_{\alpha_x}$. Hence $x \in \alpha(c)$, thus $x \in \bigcup_{\beta \in \Gamma}I_\beta$.