Answer to Question #104753 in Real Analysis for nima

Question

Let G denote the family of all open subsets of the real numbers and F the fam-
ily of all closed subsets of the real numbers. Then

Every set G in G is the union of a sequence of disjoint open intervals (called the compo-
nents of G).

Answer 1

Let $\mathcal G$ be the set of all open sets of $\mathbb R$, and $\mathfrak G$ be the set of all open intervals in $\mathbb R$.

Let $G\in\mathcal G$. Since $\mathfrak G$ is the base of topology in $\mathbb R$, we have $G=\bigcup\limits_{I\in K}I$ for some $K\subset\mathfrak G$,

Introduce a binary relation $R_0$ in $K$: $(I_1,I_2)\in R_0$ if $I_1\cap I_2\neq\varnothing$. Let $R$ be a transitive closure of $R_0$. Prove that $R$ is an equivalence relation in $K$.

Indeed,

1)It is reflexive. For every $I\in K$ we have $(I,I)\in R_0\subset R$.

2)It is transitive, because it is transitive closure of $R_0$.

3)It is symmetric. If $(I_1,I_2)\in R$, then $(I_1,I_2)\in R_0^n$ for some $n$. But since $R_0$ is symmetric, $R_0^n$ is also symmetric. So $(I_2,I_1)\in R_0^n\subset R.$

Then $R$ is an equivalence relation in $K$.

So we have $K/R$ - the factorset $K$ over $R$. Denote the class of $x\in K$ in $K/R$ by $x_R$.

We have $G=\bigcup\limits_{I\in K}I=\bigcup\limits_{x\in K}\bigcup\limits_{I\in x_R}I$.

For every $x\in K$ denote $\bigcup\limits_{I\in x_R}I$ by $[x]$ , then $G=\bigcup\limits_{x\in K}[x]$ .

Note that $[x]$ is an open set for every $x\in K$. We need to prove

1) $[x]=[y]$ or $[x]\cap [y]=\varnothing$ for every $x,y\in K$.

2) $[x]$ is an open interval in $\mathbb R$ for every $x\in K$.

Indeed,

1) $R$ is an equivalence relation, we have that $x_R=y_R$ or $x_R\cap y_R=\varnothing$ for every $x,y\in K$.

If $x_R=y_R$, then $[x]=\bigcup\limits_{I\in x_R}I=\bigcup\limits_{I\in y_R}I=[y]$.

Let $x_R\cap y_R=\varnothing$. Then for every $I\in x_R$ and $J\in y_R$ we have $(I,J)\not\in R$ (otherwise $x_R=I_R=J_R=y_R$). Since $R_0\subset R$, we have $(I,J)\not\in R_0$, that is $I\cap J=\varnothing$. So we obtain $[x]\cap [y]=\left(\bigcup\limits_{I\in x_R}I\right)\cap\left(\bigcup\limits_{J\in y_R}J\right)=\bigcup\limits_{I\in x_R}\bigcup\limits_{J\in y_R}(I\cap J)=\varnothing$.

And we proved the statement 1.

2)Take arbiarary $x\in K$. Let $u,v\in [x]$ , then there are $I,J\in x_R$ such that $u\in I$ and $v\in J$.

Since $I,J\in x_R$, we have $(I,J)\in R$, that is $(I,J)\in R_0^n$ for some $n$.

So there is $I_0,\ldots,I_n\in K$ such that $I=I_0$, $J=I_n$ and $(I_{k-1},I_k)\in R_0$ for every $k=1,\ldots,n$.

That is $I_{k-1}\cap I_k\neq\varnothing$ for every $k=1,\ldots,n$. So for every $k=1,\ldots,n$ take $u_k\in I_{k-1}\cap I_k$. Let $u_0=u$ and $u_{n+1}=v$.

We have $u_{k-1}, u_k\in I_k$. Since $I_k$ is an interval, we have $[u_{k-1},u_k]\subset I_k$, where $[u_{k-1},u_k]$ is the line segment, which connects $u_{k-1}$ and $u_k$, so $[u,v]=[u_0,u_{n+1}]\subset\bigcup\limits_{k=1}^n [u_{k-1},u_k]\subset\bigcup\limits_{k=1}^n I_k$.

But $(I,I_k)\in R_0^k\subset R$ and $(x,I)\in R$, so $I_k\in x_R$, so $[u,v]\subset\bigcup\limits_{k=1}^n I_k\subset\bigcup\limits_{I\in x_R} I=[x]$ .

So we proved that $[x]$ is an interval in $\mathbb R$. Since $[x]$ is an open set, we have that $[x]$ is an open interval.

And we proved the statement 2.

So we obtain that $G=\bigcup\limits_{x\in K}[x]$ is an union of disjoint open intervals.