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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