Semi-ring of half-closed-half-open intervals

$\newcommand{\[}{[\![}\newcommand{\]}{]\!]}\newcommand{$}{(\!(}\newcommand{$}{)\!)}$

Definition

Let $a:= ({ a_i })_{ i = 1 }^{ n }\subseteq \mathbb{R}$ ^{[Note 1]}^{[Note 2]} and $b:= ({ b_i })_{ i = 1 }^{ n }\subseteq \mathbb{R}$ be two finite sequences of the same length (namely $n\in\mathbb{N}$ ), we define $\[a,b\)$ , a half-open-half-closed rectangle in $\mathbb{R}^n$ ^[1] as follows:

$\[a,b\):=[a_1,b_1)\times\cdots\times[a_n,b_n)\subset\mathbb{R}^n$ where $[\alpha,\beta):=\{x\in\mathbb{R}\ \vert\ \alpha\le x < \beta\}$ ^{[Convention 1]}

We denote the collection of all such half-open-half-closed rectangles by $\mathscr{J}^n$ ^[1], $\mathscr{J}(\mathbb{R}^n)$ ^[1] or, provided the context makes the dimensions obvious, simply just $\mathscr{J}$ ^[1]. Formally:

$\mathscr{J}^n:=\{\[a,b\)\ \vert\ a,b\in\mathbb{R}^n\}$

Furthermore, we claim $\mathscr{J}^n$ is a semi-ring of sets

Proof of claims

[Expand]

Recall the definition of a semi-ring of sets

$\newcommand{\bigudot}{ \mathchoice{\mathop{\bigcup\mkern-15mu\cdot\mkern8mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}}{\mathop{\bigcup\mkern-13mu\cdot\mkern5mu}} }$

$\newcommand{\udot}{\cup\mkern-12.5mu\cdot\mkern6.25mu\!}$

$\require{AMScd}\newcommand{\d}[1][]{\mathrm{d}^{#1} }$

A collection of sets, $\mathcal{F}$ ^{[Note 3]} is called a semi-ring of sets if^[1]:

$\emptyset\in\mathcal{F}$
$\forall S,T\in\mathcal{F}[S\cap T\in\mathcal{F}]$
$\forall S,T\in\mathcal{F}\exists(S_i)_{i=1}^m\subseteq\mathcal{F}$ $\text{ pairwise disjoint}$ $[S-T=\bigudot_{i=1}^m S_i]$ ^{[Note 4]} - this doesn't require $S-T\in\mathcal{F}$ note, it only requires that their be a finite collection of disjoint elements whose union is $S-T$ .

In order to prove this we will first show that $\mathscr{J}^1$ (the collection of half-open-half-closed intervals of the form $[a,b)\subset\mathbb{R}$ ) is a semi-ring. Then we shall use induction on $n$ to show it for all $\mathscr{J}^n$

Grade: B

This page requires one or more proofs to be filled in, it is on a to-do list for being expanded with them.

Please note that this does not mean the content is unreliable. Unless there are any caveats mentioned below the statement comes from a reliable source. As always, Warnings and limitations will be clearly shown and possibly highlighted if very important (see template:Caution et al).
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It's a straightforward and fairly easy proof, but it isn't trivial. As such it is not marked as an easy proof, but nor does it have a high grade. See page 18 in Measures, Integrals and Martingales - René L. Schilling if stuck.

Conventions

Jump up ↑
For intervals in general we define the following:
1. If $\alpha\ge\beta$ then $[\alpha,\beta)=\emptyset$
2. If \{X_\alpha\}_{\alpha\in I} is an arbitrary collection of sets where one or more of the X_\alpha are the empty set, \emptyset, then:
  - $\prod_{\alpha\in I}X_\alpha=\emptyset$ (here $\prod$ denotes the Cartesian product)
As such if there is one or more a_i and b_i such that a_i\ge b_i then \[a,b\)=\emptyset

Notes

Jump up ↑ Or equivalently, $a\in\mathbb{R}^n$ , either way we get an $n$ -tuple of real numbers
Jump up ↑
The symbol \subset could be used instead of \subseteq but it doesn't matter, as:
- $\big[A\subset B\big]\implies\big[A\subseteq B\big]$
Jump up ↑ An F is a bit like an R with an unfinished loop and the foot at the right. "Semi Ring".
Jump up ↑
Usually the finite sequence ({ S_i })_{ i = m }^{ \infty }\subseteq \mathcal{F} being pairwise disjoint is implied by the \bigudot however here I have been explicit. To be more explicit we could say:
- \forall S,T\in\mathcal{F}\exists(S_i)_{i=1}^m\subseteq\mathcal{F}\left[\underbrace{\big(\forall i,j\in\{1,\ldots,m\}\subset\mathbb{N}[i\ne j\implies S_i\cap S_j=\emptyset]\big)}_{\text{the }S_i\text{ are pairwise disjoint} }\overbrace{\wedge}^\text{and}\left(S-T=\bigcup_{i=1}^m S_i\right)\right]
  - The statement: $\forall S,T\in\mathcal{F}\exists(S_i)_{i=1}^m\subseteq\mathcal{F}\left[\big(\forall i,j\in\{1,\ldots,m\}\subset\mathbb{N}[i\ne j\implies S_i\cap S_j=\emptyset]\big)\implies\left(S-T=\bigcup_{i=1}^m S_i\right)\right]$ is entirely different
    - In this statement we are only declaring that a finite sequence exists, and if it is NOT pairwise disjoint, then we may or may not have $S-T=\bigcup_{i=1}^mS_i$ . We require that they be pairwise disjoint AND their union be the set difference of $S$ and $T$ .

References

↑ ^{Jump up to: 1.0} ^1.1 ^1.2 ^1.3 ^1.4 Measures, Integrals and Martingales - René L. Schilling

[4] Jump up ↑ For intervals in general we define the following:
If $\alpha\ge\beta$ then $[\alpha,\beta)=\emptyset$

If $\{X_\alpha\}_{\alpha\in I}$ is an arbitrary collection of sets where one or more of the $X_\alpha$ are the empty set, $\emptyset$ , then:
$\prod_{\alpha\in I}X_\alpha=\emptyset$ (here $\prod$ denotes the Cartesian product)
As such if there is one or more $a_i$ and $b_i$ such that $a_i\ge b_i$ then $\[a,b\)=\emptyset$

[2] If $\alpha\ge\beta$ then $[\alpha,\beta)=\emptyset$

[3] If $\{X_\alpha\}_{\alpha\in I}$ is an arbitrary collection of sets where one or more of the $X_\alpha$ are the empty set, $\emptyset$ , then:
$\prod_{\alpha\in I}X_\alpha=\emptyset$ (here $\prod$ denotes the Cartesian product)

[4] $\prod_{\alpha\in I}X_\alpha=\emptyset$ (here $\prod$ denotes the Cartesian product)

[1] Jump up ↑ Or equivalently, $a\in\mathbb{R}^n$ , either way we get an $n$ -tuple of real numbers

[2] Jump up ↑ The symbol $\subset$ could be used instead of $\subseteq$ but it doesn't matter, as:
$\big[A\subset B\big]\implies\big[A\subseteq B\big]$

[7] $\big[A\subset B\big]\implies\big[A\subseteq B\big]$

[5] Jump up ↑ An F is a bit like an R with an unfinished loop and the foot at the right. "Semi Ring".

[6] Jump up ↑ Usually the finite sequence $({ S_i })_{ i = m }^{ \infty }\subseteq \mathcal{F}$ being pairwise disjoint is implied by the $\bigudot$ however here I have been explicit. To be more explicit we could say:
$\forall S,T\in\mathcal{F}\exists(S_i)_{i=1}^m\subseteq\mathcal{F}\left[\underbrace{\big(\forall i,j\in\{1,\ldots,m\}\subset\mathbb{N}[i\ne j\implies S_i\cap S_j=\emptyset]\big)}_{\text{the }S_i\text{ are pairwise disjoint} }\overbrace{\wedge}^\text{and}\left(S-T=\bigcup_{i=1}^m S_i\right)\right]$
The statement: $\forall S,T\in\mathcal{F}\exists(S_i)_{i=1}^m\subseteq\mathcal{F}\left[\big(\forall i,j\in\{1,\ldots,m\}\subset\mathbb{N}[i\ne j\implies S_i\cap S_j=\emptyset]\big)\implies\left(S-T=\bigcup_{i=1}^m S_i\right)\right]$ is entirely different
In this statement we are only declaring that a finite sequence exists, and if it is NOT pairwise disjoint, then we may or may not have $S-T=\bigcup_{i=1}^mS_i$ . We require that they be pairwise disjoint AND their union be the set difference of $S$ and $T$ .

[10] $\forall S,T\in\mathcal{F}\exists(S_i)_{i=1}^m\subseteq\mathcal{F}\left[\underbrace{\big(\forall i,j\in\{1,\ldots,m\}\subset\mathbb{N}[i\ne j\implies S_i\cap S_j=\emptyset]\big)}_{\text{the }S_i\text{ are pairwise disjoint} }\overbrace{\wedge}^\text{and}\left(S-T=\bigcup_{i=1}^m S_i\right)\right]$
The statement: $\forall S,T\in\mathcal{F}\exists(S_i)_{i=1}^m\subseteq\mathcal{F}\left[\big(\forall i,j\in\{1,\ldots,m\}\subset\mathbb{N}[i\ne j\implies S_i\cap S_j=\emptyset]\big)\implies\left(S-T=\bigcup_{i=1}^m S_i\right)\right]$ is entirely different
In this statement we are only declaring that a finite sequence exists, and if it is NOT pairwise disjoint, then we may or may not have $S-T=\bigcup_{i=1}^mS_i$ . We require that they be pairwise disjoint AND their union be the set difference of $S$ and $T$ .

[11] The statement: $\forall S,T\in\mathcal{F}\exists(S_i)_{i=1}^m\subseteq\mathcal{F}\left[\big(\forall i,j\in\{1,\ldots,m\}\subset\mathbb{N}[i\ne j\implies S_i\cap S_j=\emptyset]\big)\implies\left(S-T=\bigcup_{i=1}^m S_i\right)\right]$ is entirely different
In this statement we are only declaring that a finite sequence exists, and if it is NOT pairwise disjoint, then we may or may not have $S-T=\bigcup_{i=1}^mS_i$ . We require that they be pairwise disjoint AND their union be the set difference of $S$ and $T$ .

[12] In this statement we are only declaring that a finite sequence exists, and if it is NOT pairwise disjoint, then we may or may not have $S-T=\bigcup_{i=1}^mS_i$ . We require that they be pairwise disjoint AND their union be the set difference of $S$ and $T$ .

[MIAMRLS-3] {Jump up to: 1.0} ^1.1 ^1.2 ^1.3 ^1.4 Measures, Integrals and Martingales - René L. Schilling

[Note 1]

[Note 2]

[1]

[Convention 1]

[Note 3]

[Note 4]

Semi-ring of half-closed-half-open intervals

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Definition

Proof of claims

Conventions

Notes

References

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