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 $.


  • 3
    $\begingroup$ 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. $\endgroup$
    – desiigner
    Jul 22, 2019 at 20:05
  • 1
    $\begingroup$ 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 $x\in(a,b)\subseteq A$. $\endgroup$
    – bof
    Jul 23, 2019 at 3:06
  • $\begingroup$ @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! $\endgroup$
    – user690730
    Jul 24, 2019 at 8:53

1 Answer 1


If none of the I's are unbounded, then theorem is established.

If I is an unbounded interval, an upper ray (a,oo), then
I = $\cup${ (a,n) : n in N } expresses I as countable union of bounded intervals.
Likewise if I has no lower bound.

So every open set is a countable union of bounded I's along with a countable many open intervals needed to express to any unbounded intervals.

In the event that no I has a lower or upper bound then
A = R = $\cup${ (-n,n) : n in N } proves the theorem.


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