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Can every nonempty open set be written as a countable union of bounded open intervals of the form $(a_k,b_k)$, where $a_k$ and $b_k$ are real numbers (not $\pm\infty$)?

If yes, can someone point me toward a proof?

If not, counterexample?

Note that this is not the same question as the property "every nonempty open set is the disjoint union of a countable collection of open intervals."

David
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3 Answers3

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Hint: Let $A$ be open, and consider all intervals $(p,q)$ such that $p$ and $q$ are rational and $(p,q)\subset A$.

André Nicolas
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    OK. If $x\in\cup(p,q)$, then $x\in(p,q)$ for some rational numbers $p$ and $q$. Because $(p,q)\subset A$, then $x\in A$. Hence, $\cup(p,q)\subset A$. On the other hand, suppose $x\in A$. Because $A$ is open, there exists and $\epsilon>0$ such that $(a-\epsilon,a+\epsilon)\subset A$. Now because of the density of $Q$, there exists rational $p$ and $q$ such that $a-\epsilon<p<x<q<a+\epsilon$. Hence, $x\in\cup (p,q)$ and $A\subset\cup(p,q)$. Therefore, $\cup(p,q)=A$. Have I got it right? – David Jun 28 '13 at 22:08
  • Yes, good. It would I think be OK to consider the first containment as obvious. In principle, your argument works for the empty set too, but one might want to separate it out. – André Nicolas Jun 28 '13 at 22:16
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Yes. The answers to this question show that every non-empty open subset of $\Bbb R$ can be written as the union of countably many pairwise disjoint open intervals, when sets of the form $(a,\to)$ and $(\leftarrow,a)$ and $\Bbb R$ itself are counted as intervals. To complete the argument, you need only show that these open rays can be written as the union of countably many bounded open intervals:

$$\begin{align*} (a,\to)&=\bigcup_{n\in\Bbb Z^+}(a,a+n)\\\\ (\leftarrow,a)&=\bigcup_{n\in\Bbb Z^+}(a-n,a)\\\\ \Bbb R&=\bigcup_{n\in\Bbb Z^+}(-n,n) \end{align*}$$

Added: You can’t get pairwise disjointness in these cases, but you can arrange matters so that every point is in at most two of the bounded intervals, indeed, in the closures of at most two of them:

$$\begin{align*} (a,\to)&=\bigcup_{n\in\Bbb N}(a+2n,a+2n+3)\\\\ (\leftarrow,a)&=\bigcup_{n\in\Bbb Z^+}(a-2n-3,a-2n)\\\\ \Bbb R&=\bigcup_{n\in\Bbb Z^+}\Big((2n,2n+3)\cup(-2n-2,-2n+1)\Big)\;, \end{align*}$$

where $\Bbb N$ is the set of non-negative integers (i.e., it includes $0$).

If you don’t care about how much overlap your intervals have, of course, you can simply use the fact that the open intervals with rational endpoints are a countable base for the topology of $\Bbb R$.

Community
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Brian M. Scott
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Yes, we can.

The collection of all bounded intervals of the form $(p, q)$, where $p, q \in \Bbb Q$, form a basis for the usual topology on $\Bbb R$ (denote it by $\mathscr F$). This is easy to prove $-$ it is a common fact that the collection of all intervals (bounded or unbounded) form a basis for $\Bbb R$, say $\mathscr B$. So take any basic open interval $I \in \mathscr B$ and an element $x \in I$. If $I$ is bounded, then it is of the form $(a, b)$. Now choose two rationals $p$ and $q$ such that $ a \lt p \lt x \lt q \lt b$. Then $x \in (p, q) \in \mathscr F$.
Otherwise if it is of the form $(a,\to)$, then you can choose a natural number $n \gt x$ (archimedean property) and a rational $r$ between $a$ and $x$. Then $x \in (r, n) \in \mathscr F$. Similarly you can do for other types of unbounded intervals like $(\leftarrow,a)$ and $(-\infty, \infty)$. Hence, $\mathscr F$ is a base for the usual topology on $\Bbb R$.
Thus by the definition of a basis, every open set can be written as the union of basic open sets of $\mathscr F$ i.e. bounded open intervals (whose end points are rationals).