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 497

Example 7.4.1

An Infinite Set and a Proper Subset Can Have the Same Cardinality

Let 2\mathbb{Z} be the set of all even integers. Prove that 2\mathbb{Z} and \mathbb{Z} have the same cardinality.

Solution:

Consider the function H from \mathbb{Z} to 2\mathbb{Z} defined as follows:

 H(n) = 2n \text{ for each } n \in \mathbb{Z} 

A (partial) arrow diagram for H is shown below.

(See Page 498 for image).

To show that H is one-to-one, suppose H(n_1) = H(n_2) for some integers n_1 and n_2. Then 2n_1 = 2n_2 by definition of H, and dividing both sides by 2 gives n_1 = n_2. Hence h is one-to-one.

To show that H is onto, suppose m is any element of 2\mathbb{Z}. Then m is an even integer, and so m = 2k for some integer k. It follows that H(k) = 2k = m . Thus there exists k in \mathbb{Z} with H(k) = m, and hence H is onto.

Therefore, by definition of cardinality, \mathbb{Z} and 2\mathbb{Z} have the same cardinality.

In Section 9.4 we will show that a function from one finite set to another set of the same size is one-to-one if, and only if, it is onto. This result does not hold for infinite sets. Although it is true that for two infinite sets to have the same cardinality there must exist a function from one to the other that is both one-to-one and onto, it is always the case that:

If A and B are infinite sets with the same cardinality, then there exist functions from A to B that are one-to-one but not onto and functions from A to B that are onto but not one-to-one.

For instance, since the function H in Example 7.4.1 is one-to-one and onto, \mathbb{Z} and 2\mathbb{Z} have the same cardinality. But the "inclusion function" I from 2\mathbb{Z} to \mathbb{Z}, given by I(n) = n for all even integers n, is one-to-one but not onto. And the function J from \mathbb{Z} to 2\mathbb{Z} defined by J(n) = 2\left\lfloor \dfrac{n}{2} \right\rfloor, for each integer n, is onto but not one-to-one. (See exercise 6 at the end of this section.)

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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.