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: $$ \text{ the inverse image of } y = \{x \in X | f(x) = y\} $$ --- 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 $$ F(x) = G(x) $$ --- 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 $\log_bx$. --- 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 $f: \{0, 1\}^n \to \{0, 1\}$. --- 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 $C$**. --- 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: $$ F: X \to Y \text{ is one-to-one } \Leftrightarrow \forall x_1, x_2 \in X \text{, if } F(x_1) = F(x_2) \text{ then } x_1 = x_2 $$ --- 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. --- 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: $$ F:X \to Y \text{ is onto } \Leftrightarrow \forall y \in Y, \exists x \in X \text{ such that } F(x) = y $$ --- 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 $$ (bc)^u = b^uc^u $$ --- 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. $\log_cx = \dfrac{\log_bx}{\log_bc}$ --- 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. --- 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, $$ F^{-1}(y) = x \Leftrightarrow y = F(x) $$ --- Page 478 **Definition** The function $F^{-1}$ of Theorem 7.2.2 is called the **inverse function** for $F$. --- 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.]_ --- 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 $g$**. --- 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. --- 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. --- 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. --- 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. --- 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. --- 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. --- 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. --- **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]._ --- 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 $A$. --- 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**. --- 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]._ --- 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]._ --- 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.