Any open set $G\subset\mathbb{R}$ has the form $G=\sqcup_i (a_i,b_i)$.

Question

In real analysis class, the professor used a basic property to give an easier definition of measurable function, it says:

It is an axiom? or how to prove it?

By the way, it seems I need some background knowledge of this kind of set properties. Are there some materials I can refer?

If you want to read up about it: This statement comes from topology and is valid because the open intervals form a base of the natural topology. — raufeisen, Sep 07 '19 at 07:13
@Xuchuang just take maximal intervals in $G$. $G$ is then the disjoint union of them — mathworker21, Sep 07 '19 at 07:15
@mathworker21 Yes, this can be proved in this way, thank you. — Xuchuang, Sep 07 '19 at 08:02
https://math.stackexchange.com/questions/318299/any-open-subset-of-bbb-r-is-a-at-most-countable-union-of-disjoint-open-interv This is a big list concerning the topic. — Jan, Sep 07 '19 at 08:17

Wlod AA · Answer 1 · 2019-09-07T16:06:31.283

0

The above-given form of open sets is not canonical, the assignment of indices $i$ to the open component intervals is arbitrary. Let's write it in an equivalent but canonical way:

$\quad$ Every open set $\ G\subseteq\mathbb R\ $ is of the form $$ G\ =\ \bigcup \Gamma $$ where $\ \Gamma $ is a family of pairwise disjoint open intervals. Such family $\ \Gamma\ $ is unique and countable (possibly finite).

Indeed, define a binary relation in $\ G\ $ as follows:

$$ \forall{x,\ y\in G}\quad (\,x\equiv y\,\ \Leftrightarrow \,\ [\min(x\,\ y);\ \max(x\,\ y)]\ \subseteq G\,) $$

It's clear that

relation $\ \equiv\ $ is an equivalence relation in $G$,
every equivalence class of $\ \equiv\ $ is an open convex set,
every non-empty open convex set $\ D\subseteq\mathbb R\ $ is an open interval, namely

$$ D\ =\ (\inf D;\sup D) $$

Family $\ \Gamma\ $ is countable because of every open interval having a rational number as its point (obviously, any two different equivalence classes do not share any rational number -- after all, different classes are disjoint). .

That's all.

edited Sep 07 '19 at 16:06

answered Sep 07 '19 at 08:12

Wlod AA

2,124

Very clear. Many thanks! – Xuchuang Sep 07 '19 at 11:32
Xuchuang, thank you for good words. But, please, let me keep my notation which is mostly free of commas. I use it for years and it is accepted on the Internet almost always. It is my notation which is clearer than the standard commas which are but an eyesore. You can get used to this good thing fast. The commas are a relict of the pre-Chinese/Gutenberg times. Now we don't need them in print. – Wlod AA Sep 07 '19 at 16:04
Of course $\ (a\ b):={x\in\mathbb R: a<x<b}\ $ and $\ [a;b]:={x\in\mathbb R: a\le x\le b}\ $ etc. – Wlod AA Sep 07 '19 at 16:10
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sure, if you like. I'm OK with that. – Xuchuang Sep 08 '19 at 00:37

Any open set $G\subset\mathbb{R}$ has the form $G=\sqcup_i (a_i,b_i)$.

1 Answers1