2  Real Numbers

2.1 Fields

Definition 1: Binary operation
A binary operation on a set \(K\) is a function \[ \circ \ \colon K \times K \to K \] which maps the ordered pair \((x,y)\) into \(x \circ y\).

Definition 2

Let \(K\) be a set and \(\circ \ \colon K \times K \to K\) be a binary operation on \(K\). We say that:

  1. \(\circ\) is commutative if \[ x \circ y = y \circ x \,, \quad \, \forall \, x,y \in K \]
  2. \(\circ\) is associative if \[ (x \circ y) \circ z = x \circ (y \circ z) \,, \quad \, \forall \, x,y,z \in K \]
  3. An element \(e \in K\) is called neutral element of \(\circ\) if \[ x \circ e = e \circ x = x \,, \quad \, \forall \, x \in K \]
  4. Let \(e\) be a neutral element of \(\circ\) and let \(x \in K\). An element \(y \in K\) is called an inverse of \(x\) with respect to \(\circ\) if \[ x \circ y = y \circ x = e \,. \]
Example 3

Question. Let \(K=\{0,1\}\) be a set with binary operation \(\circ\) defined by the table \[ \begin{array}{c|cc} \circ & 0 & 1 \\ \hline 0 & 1 & 1 \\ 1 & 0 & 0 \\ \end{array} \]

  1. Is \(\circ\) commutative? Justify your answer.

  2. Is \(\circ\) associative? Justify your answer.

Solution.

  1. We have \[ 0 \circ 1 = 1 \, , \quad 1 \circ 0 = 0 \] and therefore \[ 0 \circ 1 \neq 1 \circ 0 \,. \] showing that \(\circ\) is not commutative.

  2. We have \[ (0 \circ 1) \circ 1 = 1 \circ 1 = 0 \,, \] while \[ 0 \circ (1 \circ 1) = 0 \circ 0 = 1 \,, \] so that \[ (0 \circ 1) \circ 1 \neq 0 \circ (1 \circ 1)\,. \] Thus, \(\circ\) is not associative.

Definition 4: Field

Let \(K\) be a set with binary operations of addition \[ +\ \colon K \times K \to K \,, \quad (x,y) \mapsto x + y \] and multiplication \[ \cdot\ \colon K \times K \to K \,, \quad (x,y) \mapsto x \cdot y = xy \,. \] We call the triple \((K, + , \cdot)\) a field if:

  1. The addition \(+\) satisfies: \(\,\forall \, x,y,z \in K\)
    • (A1) Commutativity and Associativity: \[ x+y = y+x \] \[ (x+y)+z = x+(y+z) \]
    • (A2) Additive Identity: There exists a neutral element in \(K\) for \(+\), which we call \(0\). It holds: \[ x + 0 = 0 + x = x \]
    • (A3) Additive Inverse: There exists an inverse of \(x\) with respect to \(+\). We call this element the additive inverse of \(x\) and denote it by \(-x\). It holds \[ x + (-x) = (-x) + x = 0 \]
  2. The multiplication \(\cdot\) satisifes: \(\,\forall \, x,y,z \in K\)
    • (M1) Commutativity and Associativity: \[ x \cdot y = y \cdot x \] \[ (x \cdot y) \cdot z = x \cdot (y \cdot z) \]
    • (M2) Multiplicative Identity: There exists a neutral element in \(K\) for \(\cdot\), which we call \(1\). It holds: \[ x \cdot 1 = 1 \cdot x = x \]
    • (M3) Multiplicative Inverse: If \(x \neq 0\) there exists an inverse of \(x\) with respect to \(\cdot\). We call this element the multiplicative inverse of \(x\) and denote it by \(x^{-1}\). It holds \[ x \cdot x^{-1} = x^{-1} \cdot x = 1 \]
  3. The operations \(+\) and \(\cdot\) are related by
    • (AM) Distributive Property: \(\,\forall \, x,y,z \in K\) \[ x \cdot (y + z) = (x \cdot y) + (y \cdot z) \,. \]
Theorem 5

Consider the sets \(\mathbb{N}\), \(\mathbb{Z}\), \(\mathbb{Q}\) with the usual operations \(+\) and \(\cdot\). We have:

  • \((\mathbb{N}, + , \cdot)\) is not a field.

  • \((\mathbb{Z}, + , \cdot)\) is not a field.

  • \((\mathbb{Q}, + , \cdot)\) is a field.

Theorem 6
Let \(K\) with \(+\) and \(\cdot\) defined by \[ \begin{array}{c|cc} + & 0 & 1 \\ \hline 0 & 0 & 1 \\ 1 & 1 & 0 \\ \end{array} \qquad \qquad \begin{array}{c|cc} \cdot & 0 & 1 \\ \hline 0 & 0 & 0 \\ 1 & 0 & 1 \\ \end{array} \] Then \((K,+,\cdot)\) is a field.

Proposition 7: Uniqueness of neutral elements and inverses

Let \((K,+,\cdot)\) be a field. Then

  1. There is a unique element in \(K\) with the property of \(0\).
  2. There is a unique element in \(K\) with the property of \(1\).
  3. For all \(x \in K\) there is a unique additive inverse \(-x\).
  4. For all \(x \in K\), \(x \neq 0\), there is a unique multiplicative inverse \(x^{-1}\).
Proof
  1. Suppose that \(0 \in K\) and \(\widetilde{0} \in K\) are both neutral element of \(+\), that is, they both satisfy (A2). Then \[ 0 + \widetilde{0} = 0 \] since \(\widetilde{0}\) is a neutral element for \(+\). Moreover \[ \widetilde{0} + 0 = \widetilde{0} \] since \(0\) is a neutral element for \(+\). By commutativity of \(+\), see property (A1), we have \[ 0 = 0 + \widetilde{0} = \widetilde{0} + 0 = \widetilde{0} \,, \] showing that \(0 = \widetilde{0}\). Hence the neutral element for \(+\) is unique.
  2. Exercise.
  3. Let \(x \in K\) and suppose that \(y, \widetilde{y} \in K\) are both additive inverses of \(x\), that is, they both satisfy (A3). Therefore \[ x + y = 0 \] since \(y\) is an additive inverse of \(x\) and \[ x + \widetilde{y} = 0 \] since \(\widetilde{y}\) is an additive inverse of \(x\). Therefore we can use commutativity and associativity and of \(+\), see property (A1), and the fact that \(0\) is the neutral element of \(+\), to infer \[\begin{align*} y & = y + 0 = y + (x + \widetilde{y}) \\ & = (y + x) + \widetilde{y} = (x + y) + \widetilde{y} \\ & = 0 + \widetilde{y} = \widetilde{y} \,, \end{align*}\] concluding that \(y = \widetilde{y}\). Thus there is a unique additive inverse of \(x\), and \[ y = \widetilde{y} = -x \,, \] with \(-x\) the element from property (A3).
  4. Exercise.
Definition 8

Let \(K\) be a set with binary operations \(+\) and \(\cdot\), and with an order relation \(\leq\). We call \((K,+,\cdot,\leq)\) an ordered field if:

  1. \((K,+,\cdot)\) is a field

  2. There \(\leq\) is of total order on \(K\): \(\, \forall \, x, y, z \in K\)

    • (O1) Reflexivity: \[ x \leq x \]
    • (O2) Antisymmetry: \[ x \leq y \, \mbox{ and } \, y \leq x \,\, \implies \,\, x = y \]
    • (O3) Transitivity: \[ x \leq y \,\, \mbox{ and } \,\, y \leq z \,\, \implies \,\, x = z \]
    • (O4) Total order:
      \[ x \leq y \,\, \mbox{ or } \,\, y \leq x \]

  3. The operations \(+\) and \(\cdot\), and the total order \(\leq\), are related by the following properties: \(\, \forall x, y, z \in K\)

    • (AM) Distributive: Relates addition and multiplication via \[ x \cdot (y + z) = x \cdot y + x \cdot z \]
    • (AO) Relates addition and order with the requirement: \[ x \leq y \,\, \implies \,\, x + z \leq y + z \]
    • (MO) Relates multiplication and order with the requirement: \[ x \geq 0, \, y \geq 0 \,\, \implies \,\, x \cdot y \geq 0 \]

Theorem 9
\((\mathbb{Q},+,\cdot,\leq)\) is an ordered field.

2.2 Supremum and infimum

Definition 10: Upper bound, bounded above, supremum, maximum

Let \(A \subseteq K\):

  1. We say that \(b \in K\) is an upper bound for \(A\) if \[ a \leq b \,, \quad \forall \, a \in A \,. \]

  2. We say that \(A\) is bounded above if there exists and upper bound \(b \in K\) for \(A\).

  3. We say that \(s \in K\) is the least upper bound or supremum of \(A\) if:

    • \(s\) is an upper bound for \(A\),
    • \(s\) is the smallest upper bound of \(A\), that is, \[ \mbox{If } \, b \in K \, \mbox{ is upper bound for } \, A \, \mbox{ then } \, s \leq b \,. \] If it exists, the supremum is denoted by \[ s = \sup \ A \,. \]

  4. Let \(A \subseteq K\). We say that \(M \in K\) is the maximum of \(A\) if: \[ M \in A \,\, \mbox{ and } \,\, a \leq M \,, \, \forall a \in A \,. \] If it exists, we denote the maximum by \[ M = \max A \,. \]

Remark 11
Note that if a set \(A \subseteq K\) in NOT bounded above, then the supremum does not exist, as there are no upper bounds of \(A\).

Proposition 12: Relationship between Max and Sup
Let \(A \subseteq K\). If the maximum of \(A\) exists, then also the supremum exists, and \[ \sup A = \max A \,. \]

Definition 13: Upper bound, bounded below, infimum, minimum

Let \(A \subseteq K\):

  1. We say that \(l \in K\) is a lower bound for \(A\) if \[ l \leq a \,, \quad \forall \, a \in A \,. \]

  2. We say that \(A\) is bounded below if there exists a lower bound \(l \in K\) for \(A\).

  3. We say that \(i \in K\) is the greatest lower bound or infimum of \(A\) if:

    • \(i\) is a lower bound for \(A\),
    • \(i\) is the largest lower bound of \(A\), that is, \[ \mbox{If } \, l \in K \, \mbox{ is a lower bound for } \, A \, \mbox{ then } \, l \leq i \,. \] If it exists, the infimum is denoted by \[ i = \inf A \,. \]

  4. We say that \(m \in K\) is the minimum of \(A\) if: \[ m \in A \,\, \mbox{ and } \,\, m \leq a \,, \, \forall a \in A \,. \] If it exists, we denote the minimum by \[ m = \min A \,. \]

Proposition 14
Let \(A \subseteq K\). If the minimum of \(A\) exists, then also the infimum exists, and \[ \inf A = \min A \,. \]

Proposition 15
Let \(A \subseteq K\). If \(\inf A\) and \(\sup A\) exist, then \[ \inf A \leq a \leq \sup A \,, \quad \forall \, a \in A \,. \]

Proposition 16: Relationship between sup and inf

Let \(A \subseteq K\). Define \[ - A := \{ - a \, \colon \,a \in A \} \,. \] They hold

  1. If \(\sup A\) exists, then \(\inf A\) exists and \[ \inf(-A) = - \sup A \,. \]
  2. If \(\inf A\) exists, then \(\sup A\) exists and \[ \sup(-A) = - \inf A \,. \]

2.3 Axioms of Real Numbers

Definition 17: Completeness

Let \((K,+,\cdot,\leq)\) be an ordered field. We say that \(K\) is complete if it holds the property:

  • (AC) For every \(A \subseteq K\) non-empty and bounded above \[ \sup A \in K \,. \]

Theorem 18
\(\mathbb{Q}\) is not complete. In particular, there exists a set \(A \subseteq \mathbb{Q}\) such that

  • \(A\) is non-empty,
  • \(A\) is bounded above,
  • \(\sup A\) does not exist in \(\mathbb{Q}\).

One of such sets is, for example, \[ A = \{ q \in \mathbb{Q}\, \colon \,q \geq 0 \,, \,\, q^2 < 2 \} \,. \]

Proposition 19
Let \((K,+,\cdot,\leq)\) be a complete ordered field. Suppose that \(A \subseteq K\) is non-empty and bounded below. Then \[ \inf A \in K\,. \]

Definition 20: System of Real Numbers \(\mathbb{R}\)

A system of Real Numbers is a set \(\mathbb{R}\) with two operations \(+\) and \(\cdot\), and a total order relation \(\leq\), such that

  • \((\mathbb{R},+,\cdot, \leq)\) is an ordered field

  • \(\mathbb{R}\) sastisfies the Axiom of Completeness

2.3.1 Inductive sets

Definition 21: Inductive set

Let \(S \subseteq \mathbb{R}\). We say that \(S\) is an inductive set if they are satisfied:

  • \(1 \in S\),
  • If \(x \in S\), then \((x + 1) \in S\).
Example 22

Question. Prove the following:

  1. \(\mathbb{R}\) is an inductive set.

  2. The set \(A=\{0,1\}\) is not an inductive set.

Solution.

  1. We have that \(1 \in \mathbb{R}\) by axiom (M2). Moreover \((x + 1) \in \mathbb{R}\) for every \(x \in \mathbb{R}\), by definition of sum \(+\).

  2. We have \(1 \in A\), but \((1 + 1) \notin A\), since \(1 + 1 \neq 0\).

Proposition 23
Let \(\mathcal{M}\) be a collection of inductive subsets of \(\mathbb{R}\). Then \[ S := \bigcap_{M \in \mathcal{M}} \, M \] is an inductive subset of \(\mathbb{R}\).

Definition 24: Set of Natural Numbers
Let \(\mathcal{M}\) be the collection of all inductive subsets of \(\mathbb{R}\). We define the set of natural numbers in \(\mathbb{R}\) as \[ \mathbb{N}:= \bigcap_{M \in \mathcal{M}} \, M \,. \]

Proposition 25: \({\mathbb{N}}_{\mathbb{R}}\) is the smallest inductive subset of \(\mathbb{R}\)
Let \(C \subseteq \mathbb{R}\) be an inductive subset. Then \[ \mathbb{N}\subseteq C \,. \] In other words, \(\mathbb{N}\) is the smallest inductive set in \(\mathbb{R}\).

Theorem 26
Let \(x \in \mathbb{N}\). Then \[ x \geq 1 \,. \]