# Covering of opens: from $\mathbb{R}$ to $\overline{\mathbb{R}}$

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!

• 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 $b$. – saulspatz Feb 28 at 16:35
• @saulspatz Yes. Thank you. And for other questions? – Jack J. Feb 28 at 17:42
• What do you mean by "the viceversa"? I don't really understand what you're trying to do: are this three steps towards... what? – Simone Ramello Mar 2 at 20:14
• @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. – Jack J. Mar 3 at 15:04
• what is the question? – 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 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 $$A\cap (b_0,a_0).$$

• 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. – zhw. Mar 6 at 18:25
• 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. – Jack J. Mar 7 at 4:03
• Disjointness in the first proof follows from the use of Thm. 1 there. I now include both proofs. – zhw. Mar 7 at 16:44