Let $A=\bigcup\limits_{i\in \mathcal I} I_i$ where $\mathcal{I}$ is an indexing set of any size.
Each $I_i$ is a nontrivial interval, i.e. an interval with at least two points. The intervals are not guaranteed to be open or closed. So any $I_i\subseteq\Bbb R$ and has the form of either $(a,b), (a,b], [a,b),$ or $[a,b]$ where $a < b, a,b\in\Bbb R$.
I want to prove that $A=\bigcup\limits_{k=1}^\infty I_{i_k}$ for some countable sub-indexing $\{i_k\}_{k=1}^\infty$.
source: exercise in Axler’s Measure, Integration & Real Analysis, section 2D, $\mathcal N\underline{o}\ 4$.
My first attempt:
We know that any union of disjoint open intervals is a countable union. Although this isn’t proved and I don’t think it’s stated in Axler’s book, it may be well-known enough to be fair game. Then, $\forall I_i$ with end-points $a_i, b_i$ we can form the set $E =\bigcup\limits_{i\in \mathcal I} \{a_i,b_i\}$ which is to say the set of all the endpoints. $\forall x_i\in E$ we take $y_i = \inf\{x\in E: x > x_i\}$ and then form the intervals $(x_i,y_i)$ which will be disjoint open intervals. (If necessary we can throw away any empty ones.)
This kind of seems like it might be heading in the right direction, but what I’m starting to notice is that I am building a collection of open intervals based on a mixture of end-points of the original intervals. I don’t see how I’m going to continue this path in order to not just get a countable union, but a countable union of the exact same intervals that were in the original set.
Edit:
The more I think about this attempt, the more I realize it's doomed in another way. If $\{I_i\}_{i\in\mathcal I}$ is the set of all intervals with left and right end-points any ordered pair of irrational numbers, then every interval constructed in the procedure above will be empty.
My next attempt is to think “Why is this problem in this section? Maybe it has a more measure theoretic solution.” If I take the measure of $A$ it might be infinite. I could consider taking the measure of $A\cap [n,n+1]$ and argue that some countable collection of intervals covers this.
However we don’t know if $A$ is closed so we can’t argue from compactness. And for all we know, even in this bounded interval, this part of $A$ might be composed of an uncountable collection of intervals.
Best Answer
Let $A$ be the union of a collection $\mathscr{C}$ of non-trivial intervals of $\mathbb{R},$ and let $B$ be the union of the interiors of the intervals in $\mathscr{C}.$
Let $P$ be the set of ordered pairs of rational numbers $\left\langle p, q \right\rangle$ such that $p < q$ and $(p, q) \subseteq I$ for some $I \in \mathscr{C},$ and choose one such interval $I = J(p, q)$ for each ordered pair $\left\langle p, q \right\rangle \in P.$ For every $x \in B,$ there exists an interval $I = (a,b), (a,b], [a,b),$ or $[a, b]$ in $\mathscr{C}$ such that $a < x < b.$ Choose rational numbers $p, q$ such that $a < p < x < q < b.$ Then $(p, q) \subseteq I \in \mathscr{C},$ therefore $\left\langle p, q \right\rangle \in P,$ and $x \in (p, q) \subseteq J(p, q).$ Therefore $B$ is contained in the union of the countable subcollection $\{J(p, q) : \left\langle p, q \right\rangle \in P\}$ of $\mathscr{C}.$
If $x \in A \setminus B,$ then $x$ is a left or right endpoint of a closed or half-closed interval belonging to $\mathscr{C}.$ Let $L$ be the set of such left endpoints, and $R$ the set of such right endpoints. (There is no assumption that $L$ and $R$ are disjoint.) For all $x \in L,$ choose an interval $H(x) = [x, h(x)] \in \mathscr{C},$ or $H(x) = [x, h(x)) \in \mathscr{C}.$ For all $x \in R,$ choose an interval $K(x) = [k(x), x] \in \mathscr{C},$ or $K(x) = (k(x), x] \in \mathscr{C}.$ For all $x, x' \in L,$ if $x \ne x'$ then $(x, h(x)) \cap (x', h(x')) = \varnothing.$ For, if $x < x',$ and $y \in (x, h(x)) \cap (x', h(x')),$ then $x' \in (x, y) \subseteq H(x),$ but this is a contradiction, because $x'$ is not an interior point of any interval in $\mathscr{C}.$ The proof is similar if $x > x'.$ Similarly, for all $x, x' \in R,$ if $x \ne x'$ then $(k(x), x) \cap (k(x'), x') = \varnothing.$ Therefore, distinct rational numbers can be chosen in $(x, h(x))$ for all $x \in L,$ and in $(k(x), x)$ for all $x \in R.$ So $L$ and $R$ are both countable.
It follows that $A$ is the union of a countable subcollection of $\mathscr{C},$ because: $$ A \subseteq \bigcup\{J(p, q) : \left\langle p, q \right\rangle \in P\} \cup \bigcup\{H(x) : x \in L\} \cup \bigcup\{K(x) : x \in R\} \subseteq A. $$