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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?

Thanks!

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    $\begingroup$ 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
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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 $A\cap (b_0,a_0).$

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  • $\begingroup$ 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
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    $\begingroup$ 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

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