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Theorem: A subset of $\mathbb{R}$ is open iff it is the union of a countable collection of open intervals.

How can I prove this theorem? I know from the properties of an open set that the union of any collection of open sets is open in $\mathbb{R}^p$, I know how to prove that, but I do not know how to prove this theorem. I know that I must use this property in my proof.

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up vote 3 down vote accepted

HINT: Let $\mathscr{B}=\{(p,q):p,q\in\Bbb Q\text{ and }p<q\}$, the set of open intervals with rational endpoints; this is a countable collection of open intervals. Prove that every open set $U$ in $\Bbb R$ is the union of the members of $B$ contained in it: $$U=\bigcup\{(p,q)\in\mathscr{B}:(p,q)\subseteq(a,b)\}\;.$$ Since $\mathscr{B}$ itself is countable, so is any subcollection of it.

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We are using $\mathbb{Q}$ because it says "union of a countable collection"? Why only $\mathbb{Q}$? – Q.matin Feb 14 '13 at 1:15
@Q.matin: Any countable dense subset of $\Bbb R$ will work: if $D$ is such a set, you can replace $\mathscr{B}$ by $$\{(p,q):p,q\in D\text{ and }p<q\}$$ and prove the same result. $\Bbb Q$ is just the handiest countable dense subset of $\Bbb R$. – Brian M. Scott Feb 14 '13 at 1:21
Thanks for that!! – Q.matin Feb 14 '13 at 1:24
@Q.matin: You’re welcome! – Brian M. Scott Feb 14 '13 at 1:25

First, prove that every open set is a union of open intervals.

Then, every open interval contains a rational representative.

Therefore, the set is a union of countably many open intervals.

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The 'rationals' answer is great, but here's another. $\mathbb{R}$ with Lebesgue measure is sigma-finite, so it suffices to show this for open subsets $U \subset (-n, n)$.

The key is that open intervals have strictly positive Lebesgue measure. Then we can use the principle that "uncountable sums are always infinite", by which I mean that if $\{a_{\alpha}\}_{\alpha \in A}$ is any set of strictly positive numbers, where $A$ is uncountable, we have

$$ \sup\{\sum_{\alpha \in E} a_{\alpha} \mid E \subset A \text{ countable}\} = \infty $$

Suppose there are uncountably many open components of $U$; let $a_{\alpha}$ denote the length of each as $\alpha$ ranges over the index set $A$ of open components of $U$. Now we use the monotonicity and countable additivity of measure to deduce that $(-n,n)$ has infinite measure, a contradiction.

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Thanks a lot !! – Q.matin Feb 15 '13 at 5:52

Use open sets of the form $(q-r,q+r)$ where $q,r\in \mathbb{Q}$ and that $\mathbb{Q}$ is dense in $\mathbb{R}$. Notice this is a countable collection. I don't really need both the $q$ and $r$ but it may help to think of $q$ as moving around $\mathbb{Q}$ and taking $r$ to be a small rational number.

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