Fields

The collection $\mathcal{F}=\left\{ A \subseteq \Omega : A \text{ is finite or } A^{c} \text{ is finite}\right\}$ is a field on $\Omega$.

Let $\mathcal{F} \subseteq 2^{\Omega }$ be such that $\Omega \in \mathcal{F}$ and $A \backslash B \in \mathcal{F}$ whenever $A, B \in \mathcal{F}$. Then $\mathcal{F}$ is a field on $\Omega$.

Every $\lambda$-system that is closed under arbitrary differences is a field.

Let $\mathcal{F} \subseteq 2^{\Omega }$ satisfy (F1) and (F2), and suppose that $\mathcal{F}$ is closed under finite disjoint unions. Then $\mathcal{F}$ is not necessarily a field.

Suppose that $\mathcal{F}_{1} \subseteq \mathcal{F}_{2} \subseteq \mathcal{F}_{3} \subseteq \ldots$, where $\mathcal{F}_{n}$ is a field on $\Omega$ for each $n \in \mathbb {N}$. Then $\bigcup_{n=1}^{\infty } \mathcal{F}_{n}$ is a field on $\Omega$.

The collection consisting of $\mathbb {R}^{k}, \varnothing$, and all $k$-dimensional rectangles of all forms fails to be a field on $\mathbb {R}^{k}$.

The collection consisting of $\varnothing$ and the finite disjoint unions of $k$-dimensional rsc subrectangles of the given $k$-dimensional rsc rectangle $(\boldsymbol {a}, \boldsymbol {b}]$ is a field on $\Omega$.

An arbitrary intersection of fields on $\Omega$ is a field on $\Omega$.

Let $\Omega$ be arbitrary, and let $\mathcal{A} \subseteq 2^{\Omega }$. There exists a unique field $\mathcal{F}$ on $\Omega$ with the properties that (i) $\mathcal{A} \subseteq \mathcal{F}$, and (ii) if $\mathcal{G}$ is a field with $\mathcal{A} \subseteq \mathcal{G}$, then $\mathcal{F} \subseteq \mathcal{G}$. This field $\mathcal{F}$ is called the [minimal] field [on $\Omega$ ] generated by A.

Let $A_{1}, \ldots , A_{n} \subsetneq \Omega$ be disjoint. What does a typical element in the minimal field generated by $\left\{ A_{1}, \ldots , A_{n}\right\}$ look like?

Let $S$ be finite, and $\Omega$ denote the set of sequences of elements of $S$. For each $\omega \in \Omega$, write

so that $z_{k}(\omega )$ denotes the $k$-th term of $\omega$ for all $k \in \mathbb {N}$. For $n \in \mathbb {N}$ and $H \subseteq S^{n}$, let

Let

Then $\mathcal{F}$ is a field of subsets of $S^{\infty }$. [The sets $C_{n}(H)$ are called cylinders of rank $n$, and $\mathcal{F}$ is collection of all cylinders of all ranks.]

Suppose that $\mathcal{A}$ is a semiring on $\Omega$ with $\Omega \in \mathcal{A}$. The collection of finite disjoint unions of $\mathcal{A}$-sets is a field on $\Omega$. [Compare with Example 3 and Exercise 1.2.7.]

Let $f: \Omega \rightarrow \Omega^{\prime }$. Given $\mathcal{A}^{\prime } \subseteq 2^{\Omega^{\prime }}$, let $f^{-1}\left(\mathcal{A}^{\prime }\right)= \left\{ f^{-1}\left(A^{\prime }\right): A^{\prime } \in \mathcal{A}^{\prime }\right\}$, where $f^{-1}\left(A^{\prime }\right)$ is the usual inverse image of $A^{\prime }$ under $f$.

Let $\Omega$ be infinite, and let $\mathcal{A} \subseteq 2^{\Omega }$ have cardinality $\aleph_{0}$. Let $f(\mathcal{A})$ denote the minimal field generated by $\mathcal{A}$ [Exercise 1.2.9]. We will show that $\operatorname {card}(f(\mathcal{A}))=\aleph_{0}$.

a. Given a collection $\mathscr {C}$, let $\mathscr {C}^{*}$ denote the collection of

1. finite unions of $\mathscr {C}$-sets,

2. finite unions of differences of $\mathscr {C}$-sets, and

3. finite unions of complements of $\mathscr {C}$-sets.

If $\varnothing \in \mathscr {C}$, then $\mathscr {C} \subseteq \mathscr {C}^{*}$. If $\operatorname {card}(\mathscr {C})=\aleph_{0}$, then $\operatorname {card}\left(\mathscr {C}^{*}\right)=\aleph_{0}$.

Some books work with a system of sets called an algebra. An algebra on $\Omega$ is a nonempty collection of subsets of $\Omega$ that satisfies (F2) and (F3).