# Introductory Real Analysis: System of Sets

Reference: Introductory Real Analysis by Kolomogorov and Fomin, Dover Pub.

Section 4:System of Sets:

Section 4.1 Rings of sets:

By a system of sets, we mean any set whose elements are themselves sets. Unless the contrary is explicitly stated, the elements of a given system of sets will be assumed to be certain subsets of some fixed set X. System of sets will be denoted by capital script letters like $\mathscr{R}, \mathscr{S}$, etc. Our chief interest will be systems of sets which have certain closure properties under some set operations.

DEFINITION 1:

A non-empty system of sets $\mathscr{R}$ is called a ring of sets if $A \triangle B \in \mathscr{R}$ and $A \bigcap B \in \mathscr{R}$ whenever $A \in \mathscr{R}$ and $B \in \mathscr{R}$.

Since

$A \bigcup B = (A \triangle B) \triangle (A \bigcap B)$,

$A - B = A \triangle (A \bigcap B)$

we also have $A \bigcup B \in mathscr{R}$ and $A - B \in \mathscr{R}$ whenever $A \in \mathscr{R}$ and $B \in \mathscr{R}$. Thus, a ring of sets is a system of sets closed under the operations of taking unions, intersections, differences, and symmetric differences. Clearly, a ring of sets is also closed under the operations of taking finite unions and intersections: $\bigcup_{k=1}^{n}A_{k}$ and $\bigcap_{k=1}^{n}A_{k}$.

A ring of sets must contain the empty set $\phi$ since $A-A=\phi$.

A set E is called the unit of a system of sets $\mathscr{S}$ if $E \in \mathscr{S}$ and $A \bigcap E =A$ for every $A \in \mathscr{S}$. Clearly, E is unique (why?). Thus, the unit of $\mathscr{S}$ is just the maximal set of $\mathscr{S}$, that is, the set containing all other sets of $\mathscr{S}$. A ring of sets with a unit is called an algebra of sets.

Example 1: Given a set A, the system $\mathscr{M}(A)$ of all subsets of A is an algebra of sets, with unit $E=A$.

Example 2: The system $\{ \phi, A\}$ consisting of the empty set $\phi$ and any nonempty set A is an algebra of sets, with $E=A$.

Example 3: The system of all finite subsets of a given set A is a ring of sets. This ring is an algebra if and only if A itself is finite.

Example 4: The system of all bounded subsets of the real line is a ring of sets, which does not contain a unit.

Theorem 1:

The intersection $\mathscr{R} = \bigcap_{\alpha} \mathscr{R}_{\alpha}$ of any set of rings is itself a ring.

Proof 1: This follows from definition 1. QED.

Theorem 2:

Given any nonempty system of sets $\mathscr{S}$, there is a unique ring $\mathscr{P}$ containing $\mathscr{S}$ and contained in every ring containing $\mathscr{S}$.

Proof 2:

If $\mathscr{P}$ exists, then clearly $\mathscr{P}$ is unique (why?). To prove the existence of $\mathscr{P}$, consider the union $X = \bigcup_{A \in \mathscr{S}}A$ of all sets A belonging to $\mathscr{S}$ and the ring $\mathscr{M}(X)$ of all subsets of X. Let $\Sigma$ be the set of all rings of sets contained in $\mathscr{M}(X)$ and containing $\mathscr{P}$. Then, the intersection $\mathscr{P}=\bigcap_{\mathscr{R} \in \Sigma}\mathscr{R}$ of all these rings clearly has the desired properties. In fact, $\mathscr{P}$ obviously contains $\mathscr{S}$. Moreover, if $\mathscr{R}^{*}$ is any ring containing $\mathscr{S}$, then the intersection $\mathscr{R}=\mathscr{R}^{*} \bigcap \mathscr{M}(X)$ is a ring in $\Sigma$ and hence, $\mathscr{P} \subset \mathscr{R} \subset \mathscr{R}^{*}$, as required. The ring $\mathscr{P}$ is called the minimal ring generated by the system $\mathscr{S}$, and will henceforth be denoted by $\mathscr{R}(\mathscr{S})$. QED.

Remarks:

The set $\mathscr{M}(X)$ containing $\mathscr{R}(\mathscr{S})$ has been introduced to avoid talking about the “set of all rings containing $\mathscr{S}$“. Such concepts as “the set of all sets,” “the set of all rings,” etc. are inherently contradictory and should be avoided. (Recall: Bertrand Russell’s famous set theory paradox).

Section 4.2:

Semirings of sets:

The following notion is more general than that of a ring of sets and plays an important role in a number of problems (especially in measure theory):

Definition 2:

A system of sets $\mathscr{S}$ is called a semiring (of sets) if

i) $\mathscr{S}$ contains the empty set $\phi$;

ii) $A \bigcap B \in \mathscr{S}$ whenever $A \in \mathscr{S}$ and $B \in \mathscr{S}$.

iii) If $\mathscr{S}$ contains the sets A and $A_{1} \in A$, then A can be represented as a finite union $A = \bigcup_{k=1}^{n}A_{k}$ …..(call this (I)) of pairwise disjoint sets of $\mathscr{S}$, with the given set $A_{1}$, as its first term.

Remarks:

The representation (I) is called a finite expansion of A, with respect to the sets $A_{1}, A_{2}, \ldots, A_{n}$.

Example 1:

Every ring of sets $\mathscr{R}$ is a semiring, since if $\mathscr{R}$ contains A and $A_{1} \subset A$, then $A = A_{1}\bigcup A_{2}$ where $A_{2}=A-A_{1} \in \mathscr{R}$.

Example 2:

The set $\mathscr{S}$ of all open intervals $(a,b)$, closed intervals $[a,b]$ and half-open intervals $[a,b)$, including the empty interval $(a,a) = \phi$ and the single element sets $[a,a] = \{ a\}$ is a semiring but not a ring.

Lemma 1:

Suppose the sets $A_{1}, A_{2}, \ldots, A_{n}$ where $A_{1}, \ldots, A_{n}$ are pairwise disjoint subsets of A, all belong to a semiring $S$. Then, there is a finite expansion

$A = \bigcup_{k=1}^{s}A_{k}$, where $s \geq n$

with $A_{1}, \ldots, A_{n}$ as its first terms, where $A_{k} \in \mathscr{S}$, $A_{k} \bigcap A_{l}=\phi$ for all $k, l = 1, \ldots, n$.

Proof of lemma 1:

The lemma holds for $n=1$ by the definition of a semiring. Suppose the lemma holds for $n=m$, and consider $m+1$ sets $A_{1}, \ldots, A_{m}, A_{m+1}$ satisfying the condition of the lemma. By hypothesis,

$A = A_{1} \bigcup \ldots \bigcup A_{m} \bigcup B_{1} \bigcup \ldots \bigcup B_{p}$ where the sets $A_{1}, \ldots, A_{m}, B_{1}, \ldots, B_{p}$ are pairwise disjoint subsets of A, all belonging to $\mathscr{S}$. Let $B_{q1}=A_{m+1} \bigcap B_{q}$

By the definition of a semiring, $B_{q} = B_{q1} \bigcup \ldots \bigcup B_{qr_{q}}$ where the sets $B_{qj} (j=1, \ldots, r_{q})$ are pairwise disjoint subsets of $B_{q}$, all belonging to $\mathscr{S}$. But then it is easy to see that

$A = A_{1}\bigcup \ldots A_{m} \bigcup A_{m+1} \bigcup \bigcup_{q=1}^{p}(\bigcup_{j=2}^{r_{q}}B_{qj})$

that is, the lemma is true for $n = m+1$. The proof now holds true by mathematical induction. QED.

Lemma 2:

Given any finite system of sets $A_{1}, \ldots, A_{n}$ belonging to a semiring $\mathscr{S}$, there is a finite system of pairwise disjoint sets $B_{1}, \ldots, B_{t}$ belonging to $\mathscr{P}$ such that every $A_{k}$ has a finite expansion $A_{k}= \bigcup_{s \in M_{k}}B_{s}$ where $k=1,2,\ldots, n$ with respect to certain of the sets $B_{s}$. (Note: Here $M_{k}$ denotes some subset of the set $\{ 1,2, \ldots, t\}$ depending on the choice of k).

Proof of Lemma 2:

The lemma is trivial for $n=1$ since we only need to set $t=1$ and $B_{1}=A_{1}$.

Let the lemma be true for $n=m$ and consider a system of sets $A_{1}, \ldots, A_{m}, A_{m+1}$ in $\mathscr{P}$. Let $B_{1}, \ldots, B_{t}$ be sets of $\mathscr{S}$ satisfying the conditions of the lemma with respect to $A_{1}, A_{2}, \ldots, A_{m}$, and let $B_{s1}=A_{m+1}\bigcap B_{s}$.

Then, by Lemma 1, there is an expansion

$A_{m+1}=(\bigcup_{s=1}^{t}B_{s1})\bigcup(\bigcup_{p=1}^{q}B_{p}^{'})$

where $B_{p}^{'} \in \mathscr{S}$. while, by definition of a semiring, there is an expansion such that

$B_{s}=B_{s1}\bigcup B_{s2} \bigcup \ldots \bigcup B_{sr}$, where $B_{sj} \in \mathscr{S}$.

It is obvious that $A_{k}= \bigcup_{s \in M_{k}}(\bigcup_{j=1}^{r_{s}}B_{sj})$ where k=1,2, …, m.

for some suitable $M_{k}$. Moreover, the sets $B_{sj}$, $B_{p}^{'}$ are pairwise disjoint. Hence, the sets $B_{sj}, B_{p}^{'}$ satisfy the conditions of the lemma with respect to $A_{1}, \ldots, A_{m}, A_{m+1}$. The proof now follows by mathematical induction. QED.

Section 4.3:

The ring generated by a semiring:

According to Theorem 1, there is a unique minimal ring $\mathscr{R}(\mathscr{S})$ generated by a system of sets $\mathscr{S}$. The actual construction of $\mathscr{R}(\mathscr{S})$ is quite complicated for arbitrary $\mathscr{S}$. However, the construction is completely straightforward if $\mathscr{S}$ is a semiring, as shown by

Theorem 3:

If $\mathscr{S}$ is a semiring, then $\mathscr{R}(\mathscr{S})$ coincides with the system $\mathscr{Z}$ of all sets A which have finite expansions

$A = \bigcup_{k=1}^{n}A_{k}$

with respect to the sets $A_{k} \in \mathscr{S}$.

Proof of Theorem 3:

First we prove that $\mathscr{Z}$ is a ring. Let A and B be any two sets in $\mathscr{Z}$. Then, there are expansions

$A = \bigcup_{i=1}^{m}A_{i}$ where $A_{i} \in \mathscr{S}$

$B = \bigcup_{j=1}^{n}B_{j}$ where $B_{i} \in \mathscr{S}$

Since $\mathscr{S}$ is a semiring, the sets $C_{ij}=A_{i} \bigcap B_{j}$ also belong to $\mathscr{S}$. By Lemma 1, there are expansions as follows:

$A_{i}=(\bigcup_{j=1}^{n}C_{ij})\bigcup(\bigcup_{k=1}^{r_{i}}D_{ik})$, where $D_{ik} \in \mathscr{S}$

$B_{j}=(\bigcup_{r=1}^{m}C_{ij})\bigcup(\bigcup_{l=1}^{s_{j}}E_{jl})$, where $E_{jl} \in \mathscr{S}$

Let us call the above two relations as (II).

It follows from (II) that $A \bigcap B$ and $A \triangle B$ have the expansions

$A \bigcap B = \bigcup_{i,j}C_{ij}$

$A \triangle B = (\bigcup_{i,k}D_{ik})\bigcup (\bigcup_{j,l}E_{jl})$.

and hence, belong to $\mathscr{Z}$. Therefore, $\mathscr{Z}$ is a ring. The fact that $\mathscr{Z}$ is a minimal ring generated by $\mathscr{S}$ is obvious. QED.

Section 4.4

Borel Algebras:

There are many problems (particularly in measure theory) involving unions and intersections not only of a finite number of sets, but also of a countable number of sets. This motivates the following concepts:

Definition 3:

$\sigma -$ ring and $\sigma -$ algebra:

A ring of sets is called a $\sigma$ -ring if it contains the union $S = \bigcup_{n=1}^{\infty}A_{n}$ whenever it contains the sets $A_{1}, A_{2}, \ldots, A_{n}, \ldots$.

A $\sigma$ -ring with a unit E is called a $\sigma$ – algebra.

Definition 4:

$\delta$ -ring and $\delta$– algebra:

A ring of sets is called a $\delta$ -ring if it contains the intersection $D = \bigcap_{n=1}^{\infty}A_{n}$

whenever it contains the sets $A_{1}, A_{2}, \ldots, A_{n}, \ldots$.

A $\delta$ -ring with a unit E is called a $\delta$ – algebra.

Theorem 4:

Every $\sigma$-algebra is a $\delta$-algebra and conversely.

Proof of theorem 4:

These are immediate consequences of the “dual” formulae:

$\bigcup_{n}A_{n}=E - \bigcap_{n}(E-A_{n})$

$\bigcap_{n}A_{n}= E -\bigcup_{\alpha}(E-A_{n})$.

QED.

The term Borel algebra or briefly B-algebra is often used to denote a $\sigma$ -algebra (equivalently, a $\delta$-algebra). The simplest example of a B-algebra is the set of all subsets of a given set A.

Given any system of sets $\mathscr{S}$, there always exists at least one B-algebra containing $\mathscr{S}$. In fact, let

$X = \bigcup_{A \in \mathscr{S}}A$

Then, the system $\mathscr{B}$ of all subsets of X is clearly a B-algebra containing $\mathscr{S}$.

If $\mathscr{B}$ is any Borel-algebra containing $\mathscr{S}$ and if E is its unit, then every $A \in \mathscr{S}$ is contained in E and hence,

$X = \bigcup_{A \in \mathscr{s}}A \subset E$.

A borel-algebra $\mathscr{B}$ is called irreducible (with respect to the system $\mathscr{S}$) if $X=E$, that is, an irreducible Borel-algebra is a Borel-algebra containing no points that do not belong to one of the sets $A \in \mathscr{S}$. In every case, it will be enough to consider only irreducible Borel-algebras.

Theorem 2 has the following analogue for irreducible Borel-algebras:

Theorem 5:

Given any non empty system of sets $\mathscr{S}$, there is a unique irreducible (to be precise, irreducible with respect to $\mathscr{S}$) B-algebra $\mathscr{B}(\mathscr{S})$ containing $\mathscr{S}$ and contained in every B-algebra containing $\mathscr{S}$.

Proof of theorem 5:

The proof is virtually identical with that of Theorem 2. The B-algebra $\mathscr{B}(\mathscr{S})$ is called the minimal B-algebra generated by the system $\mathscr{S}$ or the Borel closure of $\mathscr{S}$.

Remarks:

An important role is played in analysis by Borel sets or B-sets. These are the subsets of the real line belonging to the minimal B-algebra generated by the set of all closed intervals $[a,b]$.

Exercises to follow,

Regards,

Nalin Pithwa

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