Theorem 1. Every open subset $A\subseteq\mathbb{R}$ is countable union of disjoint open intervals. That is $$A=\bigsqcup_{q\in A\cap \mathbb{Q}} I_q.$$ Proof. For the proof I used this construction: Every open subset $A\subseteq\mathbb{R}$ is countable union of disjoint open intervals.

Theorem 2. Every open subset $A\subseteq \mathbb{R}$ is a countable union of bounded open intervals

Proof. For all $n\in\mathbb{Z}$ we consider $A_n=A\cap (n,n+2).$ Clearly $A_n$ is bounded and open (finite intersection of open) for all $n\in\mathbb{Z}$, then for the Theorem 1 $$A_n=\bigsqcup _{q\in A_n\cap\mathbb{Q}} I_q.$$ We have to show that $$A=\bigcup_{n\in\mathbb{Z}} A_n.$$ Since $A_n\subseteq A$ for all $n\in\mathbb{Z}$ we have that $\cup_{n\in\mathbb{Z}} A_n\subseteq A.$

Question 1. How can I proceed for the vice versa?

Question 2. Why in this case the open intervals are not disjoint?

Theorem 3. Every open subset $A\subset \overline{\mathbb{R}}$ is countable union of open intervals.

Proof. For this point I made use of this posts.

Question 3. Also in this case the open intervals are not disjointed? If yes, why?


  • 3
    $\begingroup$ You can't write $(0,\infty)$ as a union of disjoint, bounded open intervals. If $(a,b)\subset (0,\infty)$ there's no way for a disjoint open interval to cover $b$. $\endgroup$ – saulspatz Feb 28 at 16:35
  • $\begingroup$ @saulspatz Yes. Thank you. And for other questions? $\endgroup$ – Jack J. Feb 28 at 17:42
  • $\begingroup$ What do you mean by "the viceversa"? I don't really understand what you're trying to do: are this three steps towards... what? $\endgroup$ – Simone Ramello Mar 2 at 20:14
  • $\begingroup$ @SimoneRamello In the second step I would like to show that $$A=\bigcup_{n\in\mathbb{Z}} A_n,$$ and after I asked if in some cases we lose the fact that the intervals are disjointed. $\endgroup$ – Jack J. Mar 3 at 15:04
  • $\begingroup$ what is the question? $\endgroup$ – zhw. Mar 3 at 18:19

Question 1: Let $a\in A.$ Then $a\in (n_0,n_0+2)$ for some $n_0.$ Thus $a\in A_{n_0},$ which implies $a\in \cup A_n.$ This proves $A\subset\cup A_n$ as desired.

Here's a way to prove Thm. 2 without using Thm. 1: Let $\mathbb Q=\{q_1,q_2,\dots\}$ be the set of rationals and $\mathbb Q_+=\{r_1,r_2,\dots\}$ be the positive rationals. Define

$$E=\{(m,n))\in \mathbb N^2: (q_m-r_n, q_m+r_n)\subset A\}.$$

Then $E$ is countable and

$$A = \bigcup_{(m,n)\in E }(q_m-r_n, q_m+r_n).$$

I'll omit the proof of this for now; ask if you have questions.

Question 2: @saulspatz already pointed out that $(0,\infty)$ can not be written as a countabe union of disjoint bounded open intervals.

Question 3 (Edited): Yes, every open set $A$ in $\overline {\mathbb R}$ is the countable union of disjoint open intervals (where "open" means open in $\overline {\mathbb R}$). Previously I gave a proof by looking at cases, and applying Theorem 1. It was a little bit of a mess. I think it's easier to simply copy the proof of Theorem 1. So for each $x\in A,$ we define $I_x$ to be the largest open interval such that $x\in I_x$ and $I_x\subset A.$ (Again, "open" means open in $\overline {\mathbb R}).$ Since $A$ is open, there will be such an $I_x$ for each $x\in A.$ Thus

$$A= \bigcup_{x\in A}I_x.$$

In the same way as that for $\mathbb R,$ we now show that distinct $I_x$ are in fact disjoint. How many distinct $I_x$ can there be? Only countably many, since each $I_x$ contains a rational. That completes the proof.

Previous proof for Question 3: We do this in cases:

i) $A\subset \mathbb R$: Here we simply apply Theorem 1 to get the result.

ii) $\infty \in A,$ $-\infty \notin A.$: Then $(a,\infty]\subset A$ for some $a\in \mathbb R.$ Let $a_0= \inf \{a:(a,\infty]\subset A\}.$ If $a_0\in \mathbb R,$ then $A = (a_0,\infty] \cup (-\infty,a_0)\cap A.$ We can then apply Thm.1 to $(-\infty,a_0)\cap A$ to get the result. If $a_0=-\infty$ then $A = (-\infty,\infty]$ and there is nothing to do.

iii) $-\infty \in A,$ $\infty \notin A.$: The proof is the same as that for ii).

iv) Both $-\infty,\infty \in A.$ If $A=[-\infty,\infty],$ we're done. Otherwise define $a_0$ as in ii) and $b_0 =\sup \{b:[-\infty,b)\subset A\}.$ Both $b_0,a_0\in \mathbb R,$ $[-\infty,b_0),(a_0,\infty]\subset A,$ and $b_0\le a_0.$ If $b_0=a_0,$ then $A = [-\infty,a_0)\cup(a_0,\infty]$ and we are done. If $b_0<a_0,$ then

$$A= [-\infty,b_0)\cup(a_0,\infty]\cup A\cap (b_0,a_0)$$

and we are done by Thm.1 applied to $A\cap (b_0,a_0).$

  • $\begingroup$ I changed the proof for Question 3. It seems easier to me now. If you want, I can go back to the first proof and paste the edited version as a comment. $\endgroup$ – zhw. Mar 6 at 18:25
  • $\begingroup$ Thank you. Maybe you leave both, even if the first one is more laborious, I think it is an exercise in the light of the theorem 1. In the answer you gave I do not remember if you had justified the fact that the intervals were disjointed. $\endgroup$ – Jack J. Mar 7 at 4:03
  • 1
    $\begingroup$ Disjointness in the first proof follows from the use of Thm. 1 there. I now include both proofs. $\endgroup$ – zhw. Mar 7 at 16:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.