Check of step in proof that open subsets of the real numbers can be expressed as unions of disjoint open intervals.

Question

The following is from "An Introduction to Lebesgue Integration and Fourier Series" by Howard J. Wilcox and David L. Myers:

7.2 Theorem: Every non-empty open set $G \subset \mathbb{R}$ can be expressed uniquely as a finite or countably infinite union of pairwise disjoint open intervals.

Proof: Suppose first that $G$ is bounded. Since $G$ is open, for each $x \in G$ there is an open subinterval of $G$ containing $x$. Let $b_{x} = \mathrm{lub} \{y \mid (x,y) \subset G\}$, and $a_{x} = \mathrm{glb} \{z \mid (z,x) \subset G\}$. Let $I_{x} = (a_{x}, b_{x})$, called the component of $x$ in $G$. Clearly $x \in I_{x}$.

Now $I_{x} \subset G$, for if $w \in I_{x}$, say $x < w < b_{x}$, then by definition of $b_{x}$, there is a number $y$ such that $w < y$ and $(x,y) \subset G$. Hence $w \in G$. The case where $a_{x} < w < x$ is handled similarly. (What about $w = x$?)

Also $a_{x} \notin G$ and $b_{x} \notin G$ (see exercise 9.10).

I am attempting Exercise 9.10: Prove that $a_{x} \notin G$, in the proof of Theorem 7.2.

This is my attempt:

Suppose for contradiction that $a_{x} \in G$. Since $G$ is open, then there exists an open subinterval $(\alpha, \beta) \subset G$ containing $a_{x}$. Since $(a_{x}, x) \subset I_{x}$, and $I_{x} \subset G$, then $(a_{x}, x) \subset G$. Then since $(\alpha, a_{x}) \subset G$, and $a_{x} \in G$ by assumption, then $(\alpha, x) \subset G$. Then $\alpha \in \{z \mid (z,x) \subset G\}$. Then since $\alpha < a_{x}$, and $a_{x}$ is a lower bound of $\{z \mid (z,x) \subset G\}$, this is a contradiction.

Is this correct? Is there a simpler or more elegant proof?

Answer 1

FWIW your proof seems fine to me. It's straightforward and it efficiently uses what is necessary for the proof without anything extraneous. Personally I can't think of a significantly better approach.

Answer 2

A simple proof.

Notation. For $x,y\in \Bbb R$ let $In[x,y]=[x,y]\cup [y,x]=[\min(x,y),\max (x,y)]$.

Let $G$ be an open subset of $\Bbb R.$ For $x,y\in G$ let $x\sim y$ iff $In[x,y]\subset G.$

Obviously $\sim$ is symmetric and reflexive on $G.$ Exercise: Show that $\sim$ is transitive on G. Hint: $In[x,y]\cup In[y,z]=In[\min(x,y,z),\max(x,y,z)]$ for all $x,y,z\in \Bbb R.$

So $\sim$ is an equivalence relation on $G.$ For $x\in G$ let $[x]_{\sim}=\{y\in G:y\sim x\}.$ Exercise: $[x]_{\sim}$ is convex for each $x\in G.$

Now the set $G_{/\sim}=\{[x]_{\sim}:x\in G\}$ of $\sim$-equivalence classes is a partition of $G.$ And each $[x]_{\sim}$ is open, because if $x\in G$ then $(-r+x,r+x)\subset G$ for some $r>0,$ so $y\sim x$ for all $y \in (-r+x,r+x).$

Therefore $G_{/\sim}$ is a partition of $G$ into a family of pair-wise disjoint non-empty convex open sets.

$Any$ family $F$ of pair-wise disjoint non-empty open subsets of $\Bbb R$ is countable because $\Bbb R$ has a countable dense subset. For example, for each $f\in F$ let $\psi(f)\in f\cap \Bbb Q.$ Then $\psi:F\to \Bbb Q $ is injective, so $F$ is countable.

Therefore $G_{/\sim}$ is countable.

Remark. The notation $In[x,y]$ is ad hoc. I just found it convenient, that you don't need to distinguish the cases $x<y,x=y,$ or $x>y $.