# Every open set $A\subseteq\mathbb{R}$ is countable union of bounded open intervals: one proof.

We state first the following Theorem:

Theorem. Every open set $$A\subseteq\mathbb{R}$$ is countable union of open disjoint intervals.

Therefore if $$A\subseteq\mathbb{R}$$ is open, then $$A=\bigcup_{q\in A\cap\mathbb{Q}} I_q,$$ where $$I_q$$ is the largest open interval in $$A$$ that contains $$q$$.

I would like to show the following corollary, whose proof, after some hints given by the text, I sketched it myself. Since I don't know if it's correct, I'd be glad if any of you would take a look at it.

Corollary Every open set $$A\subseteq\mathbb{R}$$ is countable union of bounded open intervals.

Proof. For all $$n\in\mathbb{Z}$$ we consider $$A_n:=A\cap(n,n+2).$$

Intermediate question. The $$A_n$$ are not in general intervals, right?

We note that for each $$n\in\mathbb{Z}$$ the set $$A_n$$ is bounded, in fact $$A_n\subseteq (n, n+2)$$; moreover $$A_n$$ is an open of $$\mathbb{R}$$ for all $$n\in\mathbb{Z}$$, since finite intersection of open of $$\mathbb{R}.$$

Then, for the previous theorem we have that $$A_n=\bigcup_{q\in A_n\cap\mathbb{Q}} I_q,$$ where, at this point, the $$I_q$$ are bounded open intervals, because $$A_n$$ is bounded.

Claim: $$A=\bigcup_{n\in\mathbb{Z}} A_n=\bigcup_{n\in\mathbb{Z}}\bigg[\bigcup_{q\in A_n\cap\mathbb{Q}} I_q\bigg].$$

Since $$A_n\subseteq A$$ for each $$n\in\mathbb{Z}$$, then $$\bigcup_{n\in\mathbb{Z}} A_n\subseteq A.$$

Vice versa, let $$x\in A$$. As $$A$$ is open, for the theorem stated before $$A=\bigcup_{q\in A\cap\mathbb{Q}} I_q,$$ then $$x\in I_{\tilde{q}}$$, where $$\tilde{q}\in A\cap\mathbb{Q}$$.

On the other hand, $$\tilde{q}\in ([\tilde{q}], [\tilde{q}]+2),$$ therefore $$\tilde{q}\in ([\tilde{q}]:=n_0, n_0+2)\cap A.$$ Then $$\tilde{q}\in A_{n_0}\cap\mathbb{Q}.$$

hence $$x\in I_{\tilde{q}}$$ where $$\tilde{q}\in A_{n_0}\cap\mathbb{Q}$$, then $$x\in\bigcup_{q\in A_{n_0}\cup\mathbb{Q}} I_q=A_{n_0}$$, then $$x\in\bigcup_{n\in\mathbb{Z}} A_n.$$

Final questions

1. Is the proof correct?

2. Can't I see the countability of the union, that is, the countable union of a countable union is a countable union?

PS. By $$[x]$$ I mean the integer part of $$x$$.

Thanks!

• The countable union of a countable union is again countable. Think about a bijection $\mathbb{Z}\times \mathbb{Z} \leftrightarrow \mathbb{Z}$ to see this. Your proof seems correct when you wrote $A = \cup_n \cup_q I_{q,n}$ and you have proved the corollary there. Everything else is redundant I think. You are also right that the $A_n$'s need not be intervals, but instead will be countable unions of intervals. Jul 22, 2019 at 20:05
• It might be simpler to show that the collection of all intervals $(a,b)$ with $a,b\in\mathbb Q$ and $(a,b)\subseteq A$ is countable, and that the union of those intervals is the set $A$ (assuming $A$ is open). For this argument all you have to prove is that $\mathbb Q\times\mathbb Q$ is countable, and that for each point $x$ in an open set $A$ there are rational numbers $a,b$ such that $x\in(a,b)\subseteq A$.
– bof
Jul 23, 2019 at 3:06
• @desiigner You are right. it is sufficient in fact to observe that $$\bigcup_{n\in\mathbb{Z}} A_n=A\cap\bigcup_{n\in\mathbb{Z}} (n,n+2)=A\cap\mathbb{R}=A.$$ Thanks!
– user690730
Jul 24, 2019 at 8:53

I = $$\cup$${ (a,n) : n in N } expresses I as countable union of bounded intervals.
A = R = $$\cup$${ (-n,n) : n in N } proves the theorem.