# Set Theory, Relations, Functions: Preliminaries: Part VIII

SETS and FUNCTIONS:

Reference: Introductory Real Analysis by A. N. Kolmogorov and S V Fomin, Dover books.

Operations on sets:

Let A and B be any two sets. Then, by the sum or union of A and B, denoted by $A \bigcup B$, is meant the set consisting of all elements which belong to at least one of the sets A and B. More generally, by the sum or union of an arbitrary number (finite or infinite) of sets $A_{\alpha}$ (indexed by some parameter $\alpha$), we mean the set, denoted by $\bigcup_{\alpha}A_{\alpha}$ of all elements belonging to at least one of the sets $A_{\alpha}$.

By the intersection $A \bigcap B$ of two given sets A and B, we mean the set consisting of all elements which belong to both A and B. For example, the intersection of the set of all even numbers and the set of all integers divisible by 3 is the set of all integers divisible by 6. By the intersection of an arbitrary number (finite or infinite) of sets $A_{\alpha}$, we mean the set, denoted by $\bigcap_{\alpha}A_{\alpha}$ of all elements belonging to every one of the sets $A_{\alpha}$. Two sets A and B are said to be disjoint if $A \bigcap B=\phi$, that is, if they have no elements in common. More generally, let $\mathscr{F}$ be a family of sets such that $A \bigcap B=\phi$ for every pair of sets A, B in $\mathscr{F}$. Then, the sets in $\mathscr{F}$ are said to be pairwise disjoint.

It is an immediate consequence of the above definitions that the operations $\bigcup$ and $\bigcap$ are commutative and associative, that is,

$A \bigcup B=B \bigcup A$ and $(A \bigcup B) \bigcup C= A \bigcup (B \bigcup C)$; $A \bigcap B=B \bigcap A$; and, $(A \bigcap B)\bigcap C = A \bigcap (B \bigcap C)$.

Moreover, the operations $\bigcup$ and $\bigcap$ obey the following distributive laws:

$(A \bigcup B)\bigcap C = (A \bigcap C) \bigcup (B \bigcap C)$….call this I

$(A \bigcap B)\bigcup C = (A \bigcup C) \bigcap (B \bigcup C)$….call this II.

For example, suppose that $x \in (A \bigcup B)\bigcap C$, so that x belongs to the left-hand side of I. Then, x belongs to both C and $A \bigcup B$, that is, x belongs to both C and at least one of the sets A and B. But then x belongs to at least one of the sets $A \bigcap C$ and $B \bigcap C$, that is, $x \in (A \bigcap C)\bigcup (B \bigcap C)$, so that x belongs to the right hand side of I. Conversely, suppose that $x \in (A \bigcap C)\bigcup (B \bigcap C)$. Then, x belongs to at least one of the two sets $A \bigcap C$ and $B \bigcap C$. It follows that x belongs to both C and at least one of the two sets A and B, that is, $x \in C$ and $x \in A \bigcup B$, or equivalently $x \in (A \bigcup B)\bigcap C$. This proves I, and II is proved similarly.

By the difference of two sets, $A - B$ between two sets A and B (in that order), we mean the set of all elements of A which do not belong to B. Note that it is not assumed that $A \supset B$. It is sometimes convenient (e.g., in measure theory) to consider the symmetric difference of two sets A and B, denoted by $A \delta B$ and defined as the union of the two differences $A-B$ and $B-A$: $A \delta B=(A-B)\bigcup (B-A)$.

We will often be concerned later with various sets which are all subsets of some underlying basic set R, for example, various sets of points on the real line. In this case, given a set A, the difference $R-A$ is called the complement of A, denoted by $A^{'}$.

An important role is played in set theory and its applications by the following duality principle:

$R - \bigcup_{\alpha}A_{\alpha}=\bigcap_{\alpha}(R-A_{\alpha})$…call this III

$R - \bigcap_{\alpha}A_{\alpha}=\bigcup_{\alpha}(R-A_{\alpha})$…call this IV.

In words, the complement of a union equal the intersection of the complements; and, the complement of an intersection equals the union of the complements. According to the duality principle, any theorem involving a family of subsets of a fixed set R can be converted automatically into another, “dual” theorem by replacing all subsets by their complements, all unions by intersections and all intersections by unions. To prove III, suppose that $x \in R - \bigcup_{\alpha}A_{\alpha}$….call this V.

Then, x does not belong to the union $\bigcup_{\alpha}A_{\alpha}$. …call this VI.

That is, x does not belong to any of the sets $A_{\alpha}$. It follows that x belongs to each of the complements $R - \bigcup_{\alpha}A_{\alpha}$, and hence, $x \bigcap_{\alpha}(R-A_{\alpha})$….call this VII.

Conversely, suppose that VII holds, so that x belongs to every set $R - A_{\alpha}$. Then, x does not belong to any of the sets $A_{\alpha}$, that is, x does not belong to the union VI, or equivalently V holds true. This proves 3.

Homework: Prove IV similarly.

Remark:

The designation “symmetric difference” for the set $A \Delta B$ is not too apt, since $A \Delta B$ has much in common with the sum $A \bigcup B$. In fact, in $A \bigcup B$ the two statements “x belongs to A” and “x belongs to B” are joined by the conjunction “or” used in the “either …or …or both…” sense, while in $A \Delta B$ the same two statements are joined by “or” used in the ordinary “either…or…” sense (as in “to be or not to be”). In other words, x belongs to $A \bigcup B$ if and only if x belongs to either A or B or both, while x belongs to $A \Delta B$ if and only if x belongs to either A or B but not both. The set $A \Delta B$ can be regarded as a kind of “modulo-two sum” of the sets A and B, that is, a sum of the sets A and B in which elements are dropped if they are counted twice (once in A and once in B).

Functions and mappings. Images and preimages:

(Of course, we have dealt with this topic in quite detail so far in the earlier blog series; but as presented by Kolmgorov and Fomin here, the flavour is different; at any rate, it helps to revise. Revision refines the intellect.) π π π

A rule associating a unique real number $y=f(x)$ with each element of a set real numbers X is said to define a (real) function f on X. The set X is called the domain (of definition) of f, and the set Y of all numbers $f(x)$ such that $x \in X$ is called the range of f.

More generally, let M and N be two arbitrary sets. Then a rule associating a unique element $b=f(a) \in N$ with each element $a \in M$ is again said to define a function f on M (or a function f with domain M). In this more general context, f is usually called a mapping of f M into N. By the same token, f is said to map M into N(and a into b).

If a is an element of M, the corresponding element $b=f(a)$ is called the image of a (under the mapping f). Every element of M with a given element $b \in N$ as its image is called a preimage of b. Note that in general b may have several pre-images. Moreover, N may contain elements with no pre-images at all. If b has a unique pre-image, we denote this pre-image by $f^{-1}(b)$.

If A is a subset of M, the set of all elements $f(a) \in N$ such that $a \in A$ is called the image of A, denoted by $f(A)$. The set of all elements of M whose images belong to a given set $B \subset N$ is called the preimage of B, denoted by $f^{-1}(B)$. If no element of B has a preimage, then $f^{-1}(B)=\phi$. A function f is said to map M into N if $f(M)=N$, as is always the case, and onto N if $f(M)=N$. Thus, every “onto” mapping is an “into” mapping, but converse is not true.

Suppose f maps M onto N. Then, f is said to be one-to-one if each element $b \in N$ has a unique preimage $f^{-1}(b)$. In this case, f is said to establish a one-to-one correspondence between M and N, and the mapping $f^{-1}$ associating $f^{-1}(b)$ is called the inverse of f.

Theorem I:

The preimage of the union of two sets is the union of the preimages of the sets $f^{-1}(A \bigcup B)=f^{-1}(A) \bigcup f^{-1}(B)$.

Proof of Theorem I:

If $x \in f^{-1}(A \bigcup B)$, then $f(x) \in A \bigcup B$ so that $f(x)$ belongs to at least one of the sets A and B. But, then x belongs to at least one of the sets $f^{-1}(A)$ and $f^{-1}(B)$, that is, $x \in f^{-1}(A) \bigcup f^{-1}(B)$.

Conversely, if $x \in f^{-1}(A) \bigcup f^{-1}(B)$, then x belongs to at least one of the sets $f^{-1}(A)$ and $f^{-1}(B)$. Therefore, $f(x)$ belongs to at least one of the sets A and B, that is, $f(x) \in A \bigcup B$. But, then $x \in f^{-1}(A \bigcup B)$.

QED.

Theorem 2:

The preimage of the intersection of two sets is the intersection of the preimages of the sets:

$f^{-1}(A \bigcap B)=f^{-1}(A) \bigcap f^{-1}(B)$.

Proof of Theorem 2:

If $x \in f^{-1}(A \bigcap B)$, then $f(x) \in A \bigcap B$, so that $f(x) \in A$ and (meaning, simultaneously) $f(x) \in B$. But, then $x \in f^{-1}(A)$ and $x \in f^{-1}(B)$, that is, $x \in f^{-1}(A) \bigcap f^{-1}(B)$.

Conversely, if $x \in f^{-1}(A) \bigcap f^{-1}(B)$, then $x \in f^{-1}(A)$ and $x \in f^{-1}(B)$. Therefore, $f(x) \in A$ and $f(x) \in B$, that is, $f(x) \in A \bigcap B$. But, then $x \in f^{-1}(A \bigcap B)$.

QED.

Theorem 3:

The image of the union of two sets equals the union of the images of the sets $f(A \bigcup B) = f(A) \bigcup f(B)$.

Proof of theorem 3:

If $y \in f(A \bigcup B)$, then $y=f(x)$ where x belongs to at least one of the sets A and B. Therefore, $y=f(x)$ belongs to at least one of the sets $f(A)$ and $f(B)$. That is, $x \in f(A) \bigcup f(B)$.

Conversely, if $y \in f(A) \bigcup f(B)$, then $y=f(x)$. where x belongs to at least one of the sets A and B, that is, $x \in A \bigcup B$ and hence, $y=f(x) \in f(A \bigcup B)$.

QED.

Remark I:

Surprisingly enough, the image of the intersection of two sets is not so “well-behaved”. The image of the intersection of two sets does not necessarily equal the intersection of the images of the sets. For example, suppose the mapping f projects the xy-plane onto the x-axis, carrying the point $(x,y)$ into the $(x,0)$. Then, the segments $0 \leq x \leq 1$ with $y=0$, and $0 \leq x \leq 1$ with $y=1$ do not intersect, although their images coincide.

Remark 2:

In the light of above remark: consider the following: If a function is not specified on elements, it is important in general to check that f isΒ well-defined.Β That is, unambiguously defined. For example, if the set A is the union of two subsets $A_{1}$ and $A_{2}$, and we are considering a function $f: A \longrightarrow B$, then one can try to specify a function from set A to the set B $\{ 0,1\}$ by declaring that f is to map everything in $A_{1}$ to 0 and is to map everything in $A_{2}$ to 1. This unambiguously defines f unless $A_{1}$ and $A_{2}$ have elements in common in which case it is not clear whether these elements should map to 0 or to 1. Checking that this f is well-defined therefore amounts to checking that $A_{1}$ and $A_{2}$ have no intersection.

Remark 3:

Theorems 1-3 continue to hold for unions and intersections of an arbitrary number (finite or infinite) of sets $A_{\alpha}$:

$f^{-1}(\bigcup_{\alpha})A_{\alpha}=\bigcup_{\alpha}f^{-1}(A_{\alpha})$

$f^{-1}(\bigcap_{\alpha}A_{\alpha})=\bigcap_{\alpha}f^{-1}(A_{\alpha})$

$f(\bigcup_{\alpha})A_{\alpha}=\bigcup_{\alpha}f(A_{\alpha})$

Decomposition of a set into classes. Equivalence relations.

Decompositions of a given set into pairwise disjoint subsets play an important role in a great variety of problems. For example, the plane (regarded as a point set) can be decomposed into lines parallel to the x-axis, three dimensional space can be decomposed into concentric spheres, the inhabitants of a given city can be decomposed into different age groups, and so on. Any such representation of a given set M as the union of a family of pairwise disjoint subsets of M is called a decomposition or partition of M into classes.

A decomposition is usually made on the basis of a certain criterion, allowing us to assign the elements of M to one class or another. For example, the set of all triangles in the plane can be decomposed into classes of congruent triangles, or classes of triangles of equal area, the set of all functions of x can be decomposed into classes of functions all taking the same value at a given point x, and so on. Despite the great variety of such criteria, they are not completely arbitrary. For example, it is obviously impossible to partition all real numbers into classes by assigning the number b to the same class as number a if and only if $b>a$. In fact, if $b>a$, b must be assigned to same class as a but then a cannot be assigned to same class as b, since $a. Moreover, since a is not greater than itself, a cannot be assigned to the class containing itself!! As another example, it is impossible to partition the points of the plane into classes by assigning two points to the same class if and only if the distance between them is less than 1. In fact, if the distance between a and b is less than 1, and if the distance between b and c is less than 1, it does not follow that distance between a and c is less than 1. (Hint: Think of triangle inequality). Thus, by assigning a to the same class as b and b to the same class as c, we may well find that two points fall in the same class even though the distance between them is greater than 1!

These examples suggest conditions that which must be satisfied by any criterion it it is to be used as the basis for partitioning a given set into classes. Let M be a set, and let certain ordered pairs (a,b) of elements of M be called “labelled.” If $(a,b)$ is a labelled pair, we say that a is related to b by the binary relation R and we write it symbolically as aRb. For example, if a and b are real numbers, $aRb$ might mean $a, while if a and b are triangles, $aRb$ might mean that a and b have the same area. A relation between elements of M is called a relation on M, if there exists at least one labelled pair $(a,b)$ for every $a \in M$. A relation R on M is called an equivalence relation (on M) if it satisfies the following three conditions:

• Reflexivity: $aRa$ for every $a \in M$.
• Symmetry: If $aRb$, then $bRa$.
• Transititivity: If $aRb$ and $bRc$, then $aRc$.

π π π

Nalin Pithwa, more later

This site uses Akismet to reduce spam. Learn how your comment data is processed.