discrete_mathematics_with_applications/chapter_7/notes.md

Page 449

Definition

A function f from a set X to a set $Y$, denoted: f: X \to Y, is a relation from X, the domain of f, to Y, the co-domain of f, that satisfies two properties: (1) every element in X is related to some element in Y, and (2) no element in X is related to more than one element in Y. Thus, given any element x in X, there is a unique element in Y that is related to x by f. If we call this element y, then we say that "f sends x to $y$" or "f maps x to $y$" and write x \xrightarrow{f} y or f: x \to y. The unique element to which f sends x is denoted

f(x) and is called f of x, or the output of f for the input x, or the value of f at x, or the image of x under f.

The set of all values of f taken together is called the range of $f$ or the image of X under $f$. Symbolically:

 \text{range of } f = \text{ image of } X \text{ under } f = \{y \in Y | y = f(x), \text{ for some } x \text{ in } X\} 

Given an element y in Y, there may exist elements in X with y as their image. When x is an element such that f(x) = y, then x is called a preimage of $y$ or an inverse image of $y$. The set of all inverse images of y is called the inverse image of $y$. Symbolically:

 \text{ the inverse image of } y = \{x \in X | f(x) = y\} 

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Page 451

Theorem 7.1.1 A Test for Function Equality

If F: X \to Y and G: X \to Y are functions, then F = G if, and only if, F(x) = G(x) for every x \in X.

Proof:

Suppose F: X \to Y and G: X \to Y are functions; that is, F and G are relations from X to Y that satisfy the two additional function properties. Then F and G are subsets of X \times Y, and for (x, y) to be in F means that y is the unique element related to x by F, which we denote as F(x). Similarly, for (x, y) to be in G means that y is the unique element related to x by G, which we denote as G(x).

Now suppose that F(x) = G(x) for every x \in X. Then if x is any element of X,

 (x, y) \in F \Leftrightarrow y = F(x) \Leftrightarrow y = G(x) \Leftrightarrow (x, y) \in G 

because F(x) = G(x).

So F and G consist of exactly the same elements and hence F = G.

Conversely, if F = G, then for every x \in X,

 y = F(x) \Leftrightarrow (x, y) \in F \Leftrightarrow (x, y) \in G \Leftrightarrow y = G(x) 

because F and G consist of exactly the same elements.

Thus, since both F(x) and G(x) equal y, we have that

 F(x) = G(x) 

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Page 453

Definition Logarithms and Logarithmic Functions

Let b be a positive real number with b \neq 1. For each positive real number x, the logarithm with base b of $x$, written \log_bx, is the exponent to which b must be raised to obtain x. Symbolically:

 \log_bx = y \Leftrightarrow b^y = x 

The logarithmic function with base $b$ is the function from \mathbb{R}^+ to \mathbb{R} that takes each positive real number x to \log_bx.

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Page 455

Definition

An ($n$-place) Boolean function f is a function whose domain is the set of all ordered $n$-tuples of $0$'s and $1$'s and whose co-domain is the set \{0, 1\}. More formally, the domain of a Boolean function can be described as the Cartesian product of n copies of the set \{0, 1\}, which is denoted \{0, 1\^n}. Thus f: \{0, 1\}^n \to \{0, 1\}.

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Page 457

Definition

If f: X \to Y is a function and A \subseteq X and C \subseteq Y, then

 f(A) = \{y \in Y | y = f(x) \text{ for some } x \text{ in } A\} 

and

 f^{-1}(C) = \{x \in X | f(x) \in C\} 

f(A) is called the image of $A$, and f^{-1}(C) is called the inverse image of $C$.

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Page 463

Definition

Let F be a function from a set X to a set Y. F is one-to-one (or injective) if, and only if, for all elements x_1 and x_2 in X,

 \text{if } F(x_1) = F(x_2) \text{, then } x_1 = x_2 

or, equivalently,

 \text{if } x_1 \neq x_2 \text{, then } F(x_1) \neq F(x_2) 

Symbolically:

 F: X \to Y \text{ is one-to-one } \Leftrightarrow \forall x_1, x_2 \in X \text{, if } F(x_1) = F(x_2) \text{ then } x_1 = x_2 

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Page 466

Definition: Hash Function

A hash function is a function defined from a larger, possibly infinite, set of data to a smaller fixed-size set of integers.

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Page 469

Definition

Let F be a function from a set X to a set Y. F is onto (or surjective) if, and only if, given any element y in Y, it is possible to find an element x in X with the property that y = F(x).

Symbolically:

 F:X \to Y \text{ is onto } \Leftrightarrow \forall y \in Y, \exists x \in X \text{ such that } F(x) = y 

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Page 472

Laws of Exponents

If b and c are any positive real numbers and u and v are any real numbers, the following laws of exponents hold true:

7.2.1

 b^ub^v = b^{u + v} 

7.2.2

 (b^u)^v = b^{uv} 

7.2.3

 \frac{b^u}{b^v} = b^{u - v} 

7.2.4

 (bc)^u = b^uc^u 

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Page 473

Theorem 7.2.1 Properties of Logarithms

For any positive real numbers b, c, x and y with b \neq 1 and c \neq 1 and for every real number a:

a. \log_b(xy) = \log_bx + \log_by

b. \log_b\left(\dfrac{x}{y}\right) = \log_bx - \log_by

c. \log_b(x^a) = a\log_bx

d. \log_cx = \dfrac{\log_bx}{\log_bc}

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Page 475

Definition

A one-to-one correspondence (or bijection) from a set X to a set Y is a function F: X \to Y that is both one-to-one and onto.

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Page 478

Theorem 7.2.2

Suppose F: X \to Y is a one-to-one correspondence; in other words, suppose F is one-to-one and onto. Then there is a function F^{-1}: Y \to X that is defined as follows:

Given any element y in Y,

 F^{-1}(y) = \text{ that unique element } x \text{ in } X \text{ such that } F(x) \text{ equals } y 

Or, equivalently,

 F^{-1}(y) = x \Leftrightarrow y = F(x) 

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Page 478

Definition

The function F^{-1} of Theorem 7.2.2 is called the inverse function for F.

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Page 479

Theorem 7.2.3

If X and Y are sets and F: X \to Y is one-to-one and onto, then F^{-1}:Y \to X is also one-to-one and onto.

Proof:

F^{-1} is one-to-one:

Suppose y_1 and y_2 are elements of Y such that F^{-1}(y_1) = F^{-1}(y_2). [We must show that y_1 = y_2.] Let x = F^{-1}(y_1) = F^{-1}(y_2). Then x \in X, and by definition of F^{-1},

 F(x) = y_1 \text{ since } x = F^{-1}(y_1) 

and

 F(x) = y^2 \text{ since } x = F^{-1}(y_2) 

Consequently, y_1 = y_2 because each is equal to F(x). [This is what was to be shown.]

F^{-1} is onto:

Suppose x \in X. [We must show that there exists an element y in Y such that F^{-1}(y) = x.] Let y = F(x). Then y \in Y, and by definition of F^{-1}, F^{-1}(y) = x [as was to be shown.]

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Page 485

Definition

Let f: X \to Y and g: Y' \to Z be functions with the property that the range of f is a subset of the domain of g. Define a new function g \circ f: X \to Z as follows:

 (g \circ f)(x) = g(f(x)) \quad \text{ for each } x \in X 

where g \circ f is read "g circle $f$" and g(f(x)) is read "g of f of x." The function g \circ f is called the composition of f and $g$.

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Page 487

Theorem 7.3.1 Composition with an Identity Function

If f is a function from a set X to a set Y, and I_x is the identity function on X, and I_y is the identity function on Y, then

 \text{(a) } f \circ I_x = f \quad \text{ and } \quad \text{(b) } I_y \circ f = f 

Proof:

Part (a):

Suppose f is a function from a set X to a set Y and I_x is the identity function on X. Then, for each x in X,

 (f \circ I_x)(x) = f(I_x(x)) = f(x) 

Hence, by the definition of equality of functions, f \circ I_x = f, as was to be shown.

Part (b):

This is exercise 16 at the end of this section.

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Page 488

Theorem 7.3.2 Composition of a Function with Its Inverse

If f: X \to Y is a one-to-one and onto function with inverse function f^{-1}: Y \to X, then

 \text{(a) } f^{-1} \circ f = I_x \quad \text{ and } \quad \text{(b) } f \circ f^{-1} = I_y 

Proof:

Part (a):

Suppose f: X \to Y is a one-to-one and onto function with inverse function f^{-1}: Y \to X. [To show that f^{-1} \circ f = I_x, we must show that for each x \in X, (f^{-1} \circ f)(x) = x.] Let x be any element in X. Then, by definition of composition of functions,

 (f^{-1} \circ f)(x) = f^{-1}(f(x)) 

Let

 z = f^{-1}(f(x)) 

By the definition of inverse function,

 f(z) = f(x) 

and, because f is one-to-one, this implies that

 z = x 

Now z = f^{-1}(f(x)) also, and so, by substitution,

 f^{-1}(f(x)) = x 

Or, equivalently,

 (f^{-1} \circ f)(x) = x 

[as was to be shown].

Since x is any element of X and since I_x(x) = x, this proves that f^{-1} \circ f = I_x.

Part (b):

This is exercise 17 at the end of this section.

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Page 490

Theorem 7.3.3

If f: X \to Y and g: Y \to Z are both one-to-one functions, then g \circ f is one-to-one.

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Page 491

Proof of Theorem 7.3.3:

Suppose f: X \to Y and g: Y \to Z are both one-to-one functions. [We must show that g \circ f is one-to-one.] Suppose x_1 and x_2 are elements of X such that

 (g \circ f)(x_1) = (g \circ f)(x_2) 

[We must show that x_1 = x_2.] By definition of composition of functions,

 g(f(x_1)) = g(f(x_2)) 

Since g is one-to-one,

 f(x_1) = f(x_2) 

And since f is one-to-one,

 x_1 = x_2 

[as was to be shown]. Hence g \circ f is one-to-one.

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Page 491

Theorem 7.3.4

If f: X \to Y and g: Y \to Z are both onto functions, then g \circ f is onto.

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Page 493

Proof of Theorem 7.3.4

Suppose f: X \to Y and g: Y \to Z are both onto functions. [We must show that g \circ f is onto.] Let z be any [particular but arbitrarily chosen] element of Z. [We must show the existence of an element in X such that g \circ f of that element equals z.] Since g is onto, there is an element, say y, in Y such that g(y) = z. And since f is onto, there is an element, say x, in X such that f(x) = y. Hence there is an element x in X such that

 (g \circ f)(x) = g(f(x)) = g(y) = z 

[as was to be shown]. It follows that g \circ f is onto.

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Page 496

Definition

Let A and B be any sets. A has the same cardinality as $B$ if, and only if, there is a one-to-one correspondence from A to B. In other words, A has the same cardinality as B if, and only if, there is a function f from A to B that is one-to-one and onto.

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Theorem 7.4.1 Properties of Cardinality

For all sets A, B, and C:

a. Reflexive property of cardinality: A has the same cardinality as A.

b. Symmetric property of cardinality: If A has the same cardinality as B, then B has the same cardinality as A.

c. Transitive property of cardinality: If A has the same cardinality as B and B has the same cardinality as C, then A has the same cardinality as C.

Proof:

Part (a), Reflexivity:

Suppose A is any set. [To show that A has the same cardinality as A, we must show there is a one-to-one correspondence from A to A.] Consider the identity function I_A from A to A. This function is one-to-one because if x_1 and x_2 are any elements in A with I_A(x_1) = I_A(x_2), then, by definition of I_A, x_1 = x_2. The identity function is also onto because if y is any element of A, then y = I_A(y) by definition of I_A. Hence I_A is a one-to-one correspondence from A to A. [So there exists a one-to-one correspondence from A to A, as was to be shown.]

Part (b), Symmetry:

Suppose A and B are any sets and A has the same cardinality as B. [We must show that B has the same cardinality as A.] Since A has the same cardinality as B, there is a function f from A to B that is one-to-one and onto. But then, by Theorems 7.2.2 and 7.2.3, there is a function f^{-1} from B to A that is also one-to-one and onto. Hence B has the same cardinality as A [as was to be shown].

Part c, Transitivity:

Suppose A, B, and C are any sets and A has the same cardinality as B and B has the same cardinality as C. [We must show that A has the same cardinality as C.] Since A has the same cardinality as B, there is a function f from A to B that is one-to-one and onto, and since B has the same cardinality as C, there is a function g from B to C that is one-to-one and onto. But then, by Theorems 7.3.3 and 7.3.4, g \circ f is a function from A to C that is one-to-one and onto. Hence A has the same cardinality as C [as was to be shown].

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Page 497

Definition

A and B have the same cardinality if, and only if, A has the same cardinality as B or B has the same cardinality as A.

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Page 499

Definition

A set is finite if, and only if, it is the empty set or can be put into one-to-one correspondence with a set of the form \{1, 2, \dots, n\} for some positive integer n. A set is countably infinite if, and only if, it has the same cardinality as the set of positive integers \mathbb{Z}^+. A set is countable if, and only if, it is finite or countably infinite. A set that is not countable is called uncountable.

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Page 502

Theorem 7.4.2 (Cantor)

The set of all real numbers between 0 and 1 is uncountable.

Proof (by contradiction):

Suppose the set of all real numbers between 0 and 1 is countable. Then the decimal representations of these numbers can be written in a list as follows:

 0.a_{11}a_{12}a_{13}\cdots a_{1n}\cdots 

 0.a_{21}a_{22}a_{23}\cdots a_{2n}\cdots 

 0.a_{31}a_{32}a_{33}\cdots a_{3n}\cdots 

 \vdots 

 0.a_{n1}a_{n2}a_{n3}\cdots a_{nn}\cdots 

 \vdots 

[We will derive a contradiction by showing that there is a number between 0 and 1 that does not appear on this list.]

For each pair of positive integers i and j, the $j$th decimal digit of the $i$th number on the list is a_{ij}. In particular, the first decimal digit of the first number on the list is a_{11}, the second decimal digit of the second number on the list is a_{22}, and so forth. As an example, suppose the list of real numbers between 0 and 1 starts out as follows:

0. \ \boxed{2} \ 0 \ 1 \ 4 \ 8 \ 8 \ 0 \ 2 \ \dots \ 0. \ 1 \ \boxed{1} \ 6 \ 6 \ 6 \ 0 \ 2 \ 1 \ \dots \ 0. \ 0 \ 3 \ \boxed{3} \ 5 \ 3 \ 3 \ 2 \ 0 \ \dots \ 0. \ 9 \ 6 \ 7 \ \boxed{7} \ 6 \ 8 \ 0 \ 9 \ \dots \ 0. \ 0 \ 0 \ 0 \ 3 \ \boxed{1} \ 0 \ 0 \ 2 \ \dots

The diagonal elements are boxed: a_{11} is 2, a_{22} is 1, a_{33} is 3, a_{44} is 7, a_{55} is 1, and so forth.

Construct a new decimal number d = 0.d_1d_2d_3\cdots d_n \cdots as follows:

d_n = \begin{cases} 1 & \text{if } a_{nn} \neq 1 \ 2 & \text{if } a_{nn} = 1 \end{cases}

In the previous example,

d_1 \text{ is } 1 \text{ because } a_{11} = 2 \neq 1,\ d_2 \text{ is } 2 \text{ because } a_{22} = 1,\ d_3 \text{ is } 1 \text{ because } a_{33} = 3 \neq 1,\ d_4 \text{ is } 1 \text{ because } a_{44} = 7 \neq 1,\ d_5 \text{ is } 2 \text{ because } a_{55} = 1,

and so forth. Hence d would equal 0.12112\dots.

The crucial observation is that for each integer n, d differs in the $n$th decimal position from the $n$th number on the list. But this implies that d is not on the list! In other words, d is a real number between 0 and 1 that is not on the list of all real numbers between 0 and 1. This contradiction shows the falseness of the supposition that the set of all numbers between 0 and 1 is countable. Hence the set of all real numbers between 0 and 1 is uncountable [as was to be shown].

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Page 503

Theorem 7.4.3

Any subset of any countable set is countable.

Proof:

Let A be a particular but arbitrarily chosen countable set and let B be any subset of A. [We must show that B is countable.] Either B is finite or it is infinite. If B is finite, then B is countable by the definition of countable, and we are done. So suppose B is infinite. Since A is countable, the distinct elements of A can be represented as a sequence

 a_1, a_2, a_3, \dots 

Define a function g: \mathbb{Z}^+ \to B inductively as follows:

1. Search sequentially through elements of a_1, a_2, a_3, \dots until an element of B is found [This must happen eventually since B \subseteq A and B \neq \emptyset.] Call that element g(1).

2. For each integer k \geq 2, suppose g(k - 1) has been defined. Then g(k - 1) = a_i form some a_i in \{a_1, a_2, a_3, \dots\}. Starting with a_i + 1, search sequentially through a_i + 1, a_i + 2, a_i + 3, \dots trying to find an element of B. One must be found eventually because B is infinite, and \{g(1), g(2), \dots, g(k - 1)\} is a finite set. When an element of B is found, define it to be g(k).

By (1) and (2) above, the function g is defined for each positive integer.

Since the elements of a_1, a_2, a_3, \dots are all distinct, g is one-to-one. Furthermore, the searches for elements of B are sequential: Each picks up where the previous one left off. Thus every element of A is reached during some search. Moreover, all the elements of B are located somewhere in the sequence a_1, a_2, a_3, \dots, and so every element of B is eventually found and made the image of some integer. Hence g is onto. These remarks show that g is a one-to-one correspondence from \mathbb{Z}^+ to B. So B is countably infinite and thus countable [as was to be shown].

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Page 504

Corollary 7.4.4

Any set with an uncountable subset is uncountable.

Proof:

Consider the following equivalent phrasing of Theorem 7.4.3: For every set S and for every subset A of S, if S is countable, then A is countable. The contrapositive of this statement is logically equivalent to it and states: For every set S and for every subset A of S, if A is uncountable then S is uncountable. Since this is an equivalent phrasing for the corollary, the corollary is proved.