Binary Operation
Definition 3.1. A binary operation on a set $X$ is a function from $X \times X$ into $X$.
Real Number
Definition 3.2. The real numbers $\mathbb{R}$ is a set of objects satisfying Axioms 1 to 13 as listed in the following:
(A1) There is a binary operation, called addition and denoted $+$, such that $x, y \in \mathbb{R} \Longrightarrow x + y \in \mathbb{R}$,
(A2) $(x + y) + z = x + (y + z), \forall x, y, z \in \mathbb{R}$,
(A3) $x + y = y+x, \forall x, y \in \mathbb{R}$,
(A4) An additive identity exists. There exists a real number denoted $0$ which satisfies $x + 0 = 0+x = x, \forall x \in \mathbb{R}$,
(A5) Additive inverses exist. $\forall x \in \mathbb{R}, \exists y \in \mathbb{R}$ such that $x+y=0=y+x$ ($y$ is denoted $-x$ and we define $x-z$ as $x + (-z)$, $\forall x, z \in \mathbb{R}$),
(A6) There is a binary operation, called multiplication and denoted $\cdot$, such that $x, y \in \mathbb{R} \Longrightarrow x \cdot y (\text{or } xy) \in \mathbb{R}$,
(A7) $(xy)z = x(yz), \forall x, y, z \in \mathbb{R}$,
(A8) $xy = yx, \forall x, y \in \mathbb{R}$,
(A9) A multiplicative identity exists. There exists a real number, different from $0$, denoted $1$ which satisfies $x \cdot 1 = x = 1 \cdot x$,
(A10) Multiplicative inverses exist for nonzero real numbers. $\forall x \in \mathbb{R}$ with $x \neq 0$, $\exists y \in \mathbb{R}$ such that $xy = 1 = yx$, ($y$ is denoted $x^{-1}$ or $\frac{1}{x}$ and we define $\frac{x}{y} = x(y^{-1})$ if $y \neq 0$.)
(A11) $x(y+z) = xy + yz, (y+z)x = yx + zx, \forall x, y, z \in \mathbb{R}$,
(A12) There is a subset $P$ of $\mathbb{R}$ called the positive real numbers satisfying
(i) $x, y \in P \Longrightarrow x+y, xy \in P$.
(ii) If $x \in \mathbb{R}$, then exactly one of the following statements is true: $$x \in P \text{ or } x = 0 \text{ or } -x \in P$$ (A13) A nonempty subset of real numbers which is bounded above has a least upper bound.
Theorem 3.3
Theorem 3.3. The additive identity of (A4) and the additive inverse of (A5) are unique.
Proof. Let $x \in \mathbb{R}$, and suppose that $0$ and $0'$ are two additive identities. Then $0 = 0 + 0' = 0'$ by axiom 4, so the additive identity is unique. Suppose that $y$ and $z$ are two additive inverses of $x$. Then $$y = y + 0 \text{(A4)} \\ = y + (x + z) \text{(A5)} \\ = (y + x) + z \text{(A2)} \\ = 0 + z \text{(A5)} \\ = z. \text{(A4)}$$ Thus the additive inverse is unique. $\blacksquare$
Theorem 3.4
Theorem 3.4. Let $x, y, z \in \mathbb{R}$. Then the followings hold:
(1) $x \cdot 0 = 0$,
(2) $-(-x) = x$,
(3) $-(x+y)=-x-y$
(4) $xy=0 \iff x=0 \vee y=0$,
(5) $xy=xz$ and $x \neq 0 \Longrightarrow y = z$,
(6) $-(xy) = x(-y) = (-x)y$,
(7) $(-1)x = -x$.
Proof. (1) $$0 \cdot x = (0 + 0) \cdot x \text{(A4)} \\ = 0 \cdot x + 0 \cdot x \text{(A11)} \\ \Longrightarrow 0 \cdot x = 0 \text{(A4)}.$$ (2) Note that the additive inverse of $(-x)$ is $-(-x)$. Then $$x = x + 0 \text{(A4)} \\ = x + ((-x) + (-(-x))) \text{(A5)} \\ = (x + (-x)) + (-(-x)) \text{(A2)} \\ = 0 + (-(-x)) \text{(A5)} \\ = -(-x). \text{(A4)}$$ Thus $x = -(-x)$.
(3) Then $$-(x + y) = -(x+y) + 0 + 0 \text{(A4)} \\ = -(x+y) + (x -x) + (y - y) \text{(A5)} \\ = -(x+y) + (x+y) + (-x -y) \text{(A2, A3)} \\ = 0 + (-x -y) \text{(A5)} \\ = -x -y. \text{(A4)}$$ Thus $-(x+y) = -x -y$.
(4) $(\Longleftarrow)$ Clear.
$(\Longrightarrow)$ Suppose that $y \neq 0$. Then $$x = x\cdot 1 \text{(A9)} \\ x \cdot(y \cdot \frac{1}{y}) \text{(A10)} \\ = (xy) \cdot \frac{1}{y} \text{(A7)} \\ = 0.$$ Similarly, if $x \neq 0$, then $xy = 0$.
(5) $$y = y \cdot 1 \text{(A9)} \\ = y \cdot(x \cdot \frac{1}{x}) \text{(A10)} \\ = (xy) \cdot \frac{1}{x} \text{(A7, A8)} \\ = z \cdot (x \cdot \frac{1}{x}) \text{(A7, A8)} \\ = z \cdot 1 = z. \text{(A10, A9)}$$ (6) $$0 = xy - (xy) \text{(A5)} \\ = xy - (xy) + 0 \text{(A4)} \\ = xy - (xy) + x(-y) - x(-y) \text{(A5)} \\ x(y - y) - (xy) - x(-y) \text{(A3, A11)} \\ = x \cdot 0 -(xy) - x(-y) = -(xy) - x(-y) \text{(A4, A5, (1))}$$ Then $$-(xy) = -(-x(-y)) = x(-y) \text{(A5, (2))}$$ Similarly, we can show that $-(xy) = (-x)y$.
(7) $$(-1)x = -(1 \cdot x) = -x. \text{(A9), (6)} \blacksquare$$
Definition 4.1
Definition 4.1. Let $x, y \in \mathbb{R}$.
(1) $x$ is negative if $-x$ is positive,
(2) $x > y$ means $x − y$ is positive,
(3) $x ≥ y$ means $x > y$ or $x = y$,
(4) $x < y$ means $y > x$,
(5) $x ≤ y$ means $y ≥ x$.
Theorem 4.2
Theorem 4.2. Let $x, y, z \in \mathbb{R}$.
(1) $1 > 0$,
(2) $x > y \wedge y > z \Longrightarrow x > z$,
(3) $x > y \Longrightarrow x + z > y + z$,
(4) $x > y \wedge z > 0 \Longrightarrow xz > yz,$
(5) $x > y \wedge z < 0 \Longrightarrow xz < yz$
Proof. (1) By A12, $1 \in P$ or $1 = 0$ or $-1 \in P$. By A9, $1 \neq 0$. If $-1 \in P$, then $(-1) \cdot (-1) = -(-1) = 1 \in P$ by A12 and Thm 3.4. $\bigotimes$ Thus $1 = 1 - 0 \in P \iff 1 > 0$.
(2) Note that $x > y \iff x - y \in P$ and $y > z \iff y - z \in P$ by Definition 4.1. Then $(x - y) + (y - z) = x - z \in P \iff x > z$ by Definition 4.1, A2, and A4.
(3) Note that $x > y \iff x - y \in P$. Then $x - y = x - y + 0 = x - y + (z - z) = x + z - (y + z) \in P \iff x + z > y + z$ by A2, A3, A4, A5, and Definition 4.1. $\blacksquare$
Absolute Value
Definition 4.4. Let $x \in \mathbb{R}$. We define $$|x| = \begin{cases} x & \text{ if} x \geq 0 \\ -x & \text{ if} x < 0. \end{cases}$$
Theorem 4.5
Theorem 4.5. Let $x, y \in \mathbb{R}$.
(1) Let $\varepsilon > 0$. Then $|x| < \varepsilon \iff −\varepsilon < x < \varepsilon$ and $|x| ≤ \varepsilon \iff −\varepsilon ≤ x ≤ \varepsilon.$
(2) $x ≤ |x|$,
(3) $|xy| = |x||y|$,
(4) $|x + y| ≤ |x| + |y|$,
(5) $|x| - |y| \leq |x-y|$.
Proof.
Bounded
Definition 5.1. A nonempty subset $X$ of $\mathbb{R}$ is said to be bounded above (below) if there exists a real number $a$ such that $x ≤ a (x ≥ a)$ for all $x \in X$. The number $a$ is called an upper (lower) bound for $X$.
Definition 5.2. Let $X$ be a nonempty subset of $\mathbb{R}$. A number $a$ in $\mathbb{R}$ is said to be a least upper bound or a supremum for $X$ if
(i) $a$ is an upper bound for $X$.
(ii) If $b$ is an upper bound for $X$, then $a ≤ b$.
A number $a$ in $\mathbb{R}$ is said to be a greatest lower bound or an infimum for $X$ if
(i) $a$ is a lower bound for $X$.
(ii) If $b$ is a lower bound for $X$, then $b ≤ a$.
Note. We may equivalently state that if $a$ is a least upper bound for $X$ and $b < a$, then $b$ is not an upper bound for $X$, that is, if $b < a$, then $\exists x \in X$ such that $b < x$.
Theorem 5.3
Theorem 5.3. Let $X \subseteq \mathbb{R}$. Then the least upper bound for $X$ is unique.
Proof. Let $a$ and $b$ be the least upper bounds for $X$. Then $a \leq b$ because $b$ is an upper bound for $X$ and by definition 5.2. Similarly, $b \leq a$. This means that $a = b$. $\blacksquare$
Theorem 5로 우리는 least upper bound와 greatest lower bound에 대해 다음과 같은 notation을 사용할 수 있다. $$a = \sup X, \qquad b = \inf X$$
A13은 completeness axiom이라고도 불린다. 공집합이 아닌 실수 집합의 어떤 부분집합에 대해, 그 부분집합이 bounded above되어 있으면 어떤 upper bound들이 존재한다. 이때 집합의 supremum이란 upper bound들 중에 가장 작은 값인데, 여기서 upper bound들이 supremum으로 "다가간다"라는 이미지를 상상할 수 있겠는가? 다시 말해 이 axiom은 실수 집합에서 '극한'을 이야기하기 위한 공리로 생각할 수 있다. A13은 least upper bound에 관한 공리이지만, 여기서부터 greteast lower bound 버전의 statement를 증명할 수 있다.
Theorem 5.4
Theorem 5.4. A nonempty subset of real numbers which is bounded below has a greatest lower bound.
Proof. Let $X$ be a nonempty subset of $\mathbb{R}$ which is bounded below. Then there exists $y \in \mathbb{R}$ such that $y \leq x, \forall x \in X$ by definition 5.1. Let $Y$ be the set of lower bounds for $X$. Let $x \in X$. Then $y \leq x, \forall y \in Y$, which means that $Y$ is bounded above and each $x \in X$ is an upper bound for $Y$.
By A13, there exists the least upper bound $a$ for $Y$. Then $a \leq x, \forall x \in X$ by definition 5.2, that is, $a$ is a lower bound for $X$. Furthermore, $y \leq a, \forall y \in Y$ because $a$ is an upper bound for $Y$. By definition 5.2, $a$ is the greatest lower bound for $X$. $\blacksquare$
Exercise 5.1
Exercise 5.1. Let $X$ be a set of real numbers with least upper bound $a$. Prove that if $\varepsilon > 0$, there exists $x \in X$ such that $a - \varepsilon < x \leq a$.
Solution. Note that $a - \varepsilon < a$ and $a$ is the least upper bound for $X$, so $a - \varepsilon$ is not an upper bound for $X$. Then $\exists x \in X$ such that $a - \varepsilon < x \leq a$ by definition 5.1. $\blacksquare$