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In my book's exercises section I am asked to prove that every bounded, open set in $\mathbb{R}$ is the union of disjoin open intervals.

Looking around the internet I have found many strategies that involve assigning each interval to a rational number. This has the effect of not only showing that the claim holds, but also showing that the intervals being union-ed are countable. As far as I have seen, they mostly proceed like this:

Figure that any member of a nonempty open subset $A$ of $\mathbb{R}$ is part of some open interval within $A$. By properties of intervals in $\mathbb{R}$, that interval contains a rational. With this reasoning, then, we can get an open interval corresponding to any point in $A$. Choose only the disjoint intervals. Since $\mathbb{Q}$ is countable, the number of intervals is also countable. The union of these disjoint intervals will be $A$.

Is there another way to do this that doesn't involve an extraneous set like $\mathbb{Q}$ and doesn't end up proving a stronger claim? More specifically, is there a way to solve it that depends on properties of bounded open sets (and possibly balls)?

If not, how would you deduce from the desired outcome that you would need to involve the subset $\mathbb{Q}$ ? It seems to come out of nowhere that you need to use its density properties in $\mathbb{R}$ to complete the proof.

Thank you!

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9  
But $\mathbb Q$ is not extraneous to $\mathbb R$. It is... intraneous. :-) –  Asaf Karagila Dec 20 '12 at 23:32
3  
I can't make any sense of your argument. I also can't think of any sensible way to answer this question that does use $\Bbb{Q}$. $\Bbb{Q}$ is often used to prove every open set in $\Bbb{R}$ is a union of countably many disjoint open intervals, but that's another question. –  Chris Eagle Dec 21 '12 at 0:11
    
@ChrisEagle Right; the proofs I saw all show a stronger claim that the union of intervals not only compose the set, but also that they are countable. I was wondering if there was a way that does not do this. I'll edit the question to reflect this. –  enthdegree Dec 21 '12 at 2:49

3 Answers 3

up vote 3 down vote accepted

Let $U$ be a bounded, non-empty open subset of $\Bbb R$. For $x,y\in\Bbb R$ let

$$I(x,y)=\begin{cases} [x,y],&\text{if }x\le y\\ [y,x],&\text{if }y\le x\;. \end{cases}$$

$I(x,y)$ is just the closed interval with endpoints $x$ and $y$, irrespective of whether $x\le y$ or $y\le x$.

Define a relation $\sim$ on $U$ as follows: for $x,y\in U$, $x\sim y$ iff $I(x,y)\subseteq U$. It’s easy to check that $\sim$ is an equivalence relation. Let $J$ be a $\sim$-equivalence class. $J$ is bounded, so let $a=\inf J$ and $b=\sup J$. Note that $I(x,y)\subseteq J$ for any $x,y\in J$, so $J$ has no ‘holes’ and must be one of the four intervals with endpoints $a$ and $b$, $(a,b),[a,b),(a,b]$, or $[a,b]$.

Suppose that $a\in J$; then $a\in U$, so there is an open interval $(a-\epsilon,a+\epsilon)\subseteq U$. Let $u=a-\frac{\epsilon}2$; clearly $u\sim a\in J$, so $u\in J$, contradicting the choice of $a$. Thus, $a\notin J$. A similar argument shows that $b\notin J$. Thus, we must have $J=(a,b)$

That is, each of the $\sim$-equivalence classes is an open interval. The equivalence classes of any equivalence relation partition the underlying set of that relation, so the $\sim$-equivalence classes partition $U$ into disjoint open intervals.

The same argument works even if $U$ is unbounded, though in that cases one or two of the equivalence classes can be an unbounded open interval: if $U=\Bbb R$ there is just one, $\Bbb R$ itself, and otherwise you can get one interval of the form $(\leftarrow,a)$ or one interval of the form $(a,\to)$ or both.

You need $\Bbb Q$ only to show that this decomposition has only countably many pieces. And you do need $\Bbb Q$ or some other countable dense subset of $\Bbb R$ for that: the argument that I used above actually shows that in every linearly ordered space every open set can be written as a union of disjoint intervals, but there are non-separable linear orders in which some open sets have only decompositions with uncountably many pieces.

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Suppose the open set is called $S$. For each $x \in S$ let $${\mathcal{I}}_{x} = \{ I \subseteq S \colon I \text{ is an open interval and } x \in I \} .$$ Then $\cup {\mathcal{I}}_{x}$ is an open interval containing $x$. Notice that

  • $\cup {\mathcal{I}}_{x} \subseteq S$.
  • $\cup {\mathcal{I}}_{x}$ is an interval. In fact it is the largest interval that contains $x$ and is also a subset of $S$.

Suppose that $x, y \in S$. If $\cup {\mathcal{I}}_{x}$ and $\cup {\mathcal{I}}_{y}$ have nonempty intersection then the two sets are equal. (The union of intervals that intersect is an interval.} But then $S$ is the union of disjoint intervals.

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For each $x \in S$, fix some $I_x$ such that $x \in I_x \subset S$. The union of these intervals equals $S$. If any intervals overlap, let $J$ be an index set defining the maximal overlap. Then $I_x \subset S$  for all $x \in J$, $\bigcup_{x \in J} I_x \subset S$, an open interval by construction.  

Let $y := \inf ( x, x \in J)$, $y$ not in $S$, $y \geq a$ for some $a$ by boundedness, and denote $K_y := \bigcup_{x \in J} I_x$. Define such an open interval $K_y$ for any part of $S$ where overlaps occur, and note that by construction $K_y \cap K_z = \emptyset$ when $y, z$ were defined by this procedure, and $z > y$ (say). Then this procedure yields an index set $K$ such that   $$\bigcup_{y \in K} K_y = \bigcup_{x \in S} I_x =S,$$   is a disjoint union.

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One should note that by boundedness from above, all $y$ resulting from this procedure are finite. –  gnometorule Dec 21 '12 at 0:22
    
What do you mean by "let $J$ be an index set defining the maximal overlap"? –  Potato Dec 21 '12 at 0:41
    
For any $x$, all $y$ such that $I_x$ intersects $I_y$. –  gnometorule Dec 21 '12 at 0:45

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