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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.

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    $\begingroup$ The full name of the book from which this exercise comes is Measure, Integration & Real Analysis. The electronic version of the book is legally free to the world at measure.axler.net. $\endgroup$ Jun 13 '20 at 4:08
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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: $$ A \subseteq \bigcup\{J(p, q) : \left\langle p, q \right\rangle \in P\} \cup \bigcup\{H(x) : x \in L\} \cup \bigcup\{K(x) : x \in R\} \subseteq A. $$

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    $\begingroup$ To be clear, $J(p,q)$ is just notation for the interval containing the open interval $(p,q)$? That is to say $J:P\rightarrow \mathcal{C}$. $\endgroup$
    – Addem
    Jun 13 '20 at 17:31
  • $\begingroup$ Yes, exactly. This is one of five (!) places where I appeal to the Axiom of Choice. (I can't escape the feeling that there must be a much simpler proof than either of the answers posted so far.) $\endgroup$ Jun 13 '20 at 17:52
  • $\begingroup$ I have to agree, and yet I feel much better about the answers given being so difficult. I might not be bright but at least that's not the reason for me failing to get this particular solution! It's just a hard to figure out how to make the solution easy. :) $\endgroup$
    – Addem
    Jun 13 '20 at 17:58
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    $\begingroup$ By the way, I'm still trying to process this solution, but I think when you write $I\in\mathscr{S}$ you may have meant $I\in \mathscr{C}$. $\endgroup$
    – Addem
    Jun 13 '20 at 18:26
  • $\begingroup$ No worries and thanks for contributing in the midst of so many things! I've read to the end and don't see any other typos. $\endgroup$
    – Addem
    Jun 13 '20 at 18:52
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The theorem that every open subset of $\mathbb{R}$ is a union of countably many open intervals is proposition 0.59 in Axler's Supplement for Measure, Integration & Real Analysis, so may be presumed known.

As before, 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 $\mathscr{C}.$ Because $B$ is open, it is a union of countably many open intervals. Each such interval is a union of countably many bounded closed intervals. Each of these is compact, therefore covered by the interiors of finitely many intervals in $\mathscr{C}.$ Therefore $B$ is covered by countably many intervals in $\mathscr{C}.$

Define $D = \bigcup\{I \setminus \operatorname{Int} I : I \in \mathscr{C}\}.$ Then $A = D \cup B = (D \setminus B) \cup B.$ If $x \in D$ then an interval $(x, y)$ or $(y, x)$ is contained in $B,$ therefore every element of $D \setminus B$ is an endpoint of a component interval of $B$; therefore $D \setminus B$ is countable. By the definition of $D,$ each point of $D \setminus B$ is contained in an interval in $\mathscr{C}$; choose one arbitrarily. With the cover of $B,$ this expresses $A$ as a union of countably many intervals in $\mathscr{C}.$

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Definition: For $x,y \in \Bbb R$ let $In[x,y]=[x,y]\cup [y,x].$ That is, $In[x,y]$ is the closed interval with end-point(s) $x,y,$ regardless of whether $x<y$ or $x>y$ or $x=y.$

For $x,y\in A$ let $x\sim y$ iff $In[x,y]\subseteq \cup_{i\in J}I_i$ for some countable $J\subseteq \mathcal I.$

The binary relation $\sim$ on $A$ is (obviously) reflexive and symmetric. To show that $(x\sim y\land y\sim z)\implies x\sim z$ without a list of cases (such as $x<y<z$ or $x<z<y$ etc.), observe that $In[x,y]\cup In [y,z]=[\min(x,y,z),\max(x,y,z)],$ so use $In[x,z]\subseteq [\min(x,y,z),\max(x,y,z)]=In[x,y] \cup In[y,z].$

So $\sim$ is an equivalence relation on $A.$ For $x\in A$ let $[x]_{\sim}=\{y:x\sim y\}.$ Let $B=A_{\sim}=\{[x]_{\sim}:x\in A\}.$

For $b\in B$ let $F(b)\in \Bbb Q\cap b.$ Equivalence classes are pair-wise disjoint so $F$ is injective so $B$ is countable.

For $b\in B$ take some $x\in b.$ Let $\{y_n:n\in \Bbb N\}\subseteq b\cap [x,\infty)$ such that $b\cap [x,\infty)=\cup_{n\in \Bbb N}[x,y_n].$ It does not matter whether $y_n=y_m$ for some $n\ne m.$ And if $\max b$ exists let $y_1=\max b.$

Now each $[x,y_n]\subseteq \cup_{i\in I_n}I_i$ for some countable $I_n\subseteq \mathcal I,$ so $b\cap [x,\infty)\subseteq \cup_{i\in b^+}I_i$ where $b^+=\cup_{n\in \Bbb N}I_n$ is countable.

Similarly $b\cap (-\infty,x]\subset \cup_{i\in b^-}I_i$ for some countable $b^-\subseteq \mathcal I.$

Remark: It turns out that $\sim$ is merely a means to an end. When we are done, we see that $B=\{A\},$ that is, $x\sim y$ for all $x,y\in A.$

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  • $\begingroup$ In the transitivity claim, this may have been obvious, but I'll just note that here we have one countable collection for $x\sim y$ and another for $y\sim z$. The fact that $x\sim z$ will then be exhibited by the union of those two countable collections. $\endgroup$
    – Addem
    Jun 13 '20 at 7:37
  • $\begingroup$ Yes. Obviousness is subjective. $\endgroup$ Jun 13 '20 at 15:05

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