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Section 3.3 Ordered fields

Definition 3.3.1. Total orderings.

Let \(S\) be a set. A total ordering \(\lt\) on \(S\) is a relation \(\lt\) thereon satisfying the following axioms:

  1. Transitivity. For every \(x,y,z\in S\text{,}\) if \(x\lt y\) and \(y\lt z\text{,}\) then \(x\lt z\text{.}\)

  2. Trichotomy. For every \(x,y \in S\text{,}\) exactly one of the following is true: \(x=y\text{,}\) \(x\lt y\text{,}\) or \(x\gt y\text{.}\)

A totally ordered set is a pair \((S, \lt)\) consisting of a set \(S\) along with a total ordering \(\lt\text{.}\) In this case, let us write \(x \leq y\) if and only if either \(x=y\) or else \(x \lt y\text{.}\)

Definition 3.3.2. Ordered fields.

An ordered field \(F\) is a field along with a total ordering \(\lt\) satisfying the following axioms:

  1. Transitivity. For every \(x,y,z\in F\text{,}\) if \(x\lt y\) and \(y\lt z\text{,}\) then \(x\lt z\text{.}\)

  2. Trichotomy. For every \(x,y \in F\text{,}\) exactly one of the following is true: \(x=y\text{,}\) \(x\lt y\text{,}\) or \(x\gt y\text{.}\)

  3. Additive compatibility with order. For every \(x,y,z\in F\text{,}\) if \(x\lt y\text{,}\) then \(x+z=y+z\text{.}\)

  4. Multiplicative compatibility with order. For every \(x,y\in F\text{,}\) if \(0\lt x\) and \(0\lt y\text{,}\) then \(0\lt xy\text{.}\)

Let \(m/n, p/q \in \QQ\text{.}\) (Recall that \(n,q\in\NN\text{.}\)) Then we declare that \(m/n \lt p/q\) if and only if \(mq \lt np\text{.}\)

Definition 3.3.4. Infimum and supremum.

Let \((S,\lt)\) be a totally ordered set, and let \(T \subseteq S\) be a subset.

A lower bound in \(S\) for \(T\) is an element \(x \in S\) such that for every \(t \in T\text{,}\) one has \(x \leq t\text{;}\) dually, an upper bound in \(S\) for \(T\) is an element \(y \in S\) such that for every \(t \in T\text{,}\) one has \(t \leq y\text{.}\)

An infimum or greatest lower bound in \(S\) for \(T\) is a lower bound \(x \in S \) for \(T\) such that for every lower bound \(x'\in S\) for \(T\text{,}\) one has \(x' \leq x\text{.}\) We shall write \(x=\inf T\text{.}\)

A supremum or least upper bound in \(S\) for \(T\) is an upper bound \(y \in S \) for \(T\) such that for every upper bound \(y'\in S\) for \(T\text{,}\) one has \(y \leq y'\text{.}\) We shall write \(y=\sup T\text{.}\)

If an infinmum in \(S\) for \(T\) exists, then it is unique. Indeed, if both \(x_1,x_2\in S\) are infima of \(T\text{,}\) then they are both lower bounds. But then by definition, both \(x_1 \leq x_2\) and \(x_2 \leq x_1\text{.}\)

  • Consider the set \(T = \{x \in \QQ : x \lt 0\}\text{.}\) Then \(T\) has no lower bound in \(\QQ\text{,}\) but it does have an upper bound and in fact a supremum: \(0 = \sup T\text{.}\) Note that in this case, \(\sup T \notin T\text{.}\)

  • Consider the set \(T = \{x \in \QQ : x \leq 0\}\text{.}\) Then \(T\) has no lower bound in \(\QQ\text{,}\) but it does have an upper bound and in fact a supremum: \(0 = \sup T\text{.}\) Note that in this case, \(\sup T \in T\text{.}\)

  • Consider the set \(T = \{x \in \QQ : x \gt 0\}\text{.}\) Then \(T\) has no upper bound in \(\QQ\text{,}\) but it does have a lower bound and in fact an infimum: \(0 = \inf T\text{.}\) Note that in this case, \(\inf T \notin T\text{.}\)

  • Consider the set \(T = \{x \in \QQ : x \geq 0\}\text{.}\) Then \(T\) has no upper bound in \(\QQ\text{,}\) but it does have a lower bound and in fact an infimum: \(0 = \inf T\text{.}\) Note that in this case, \(\inf T \in T\text{.}\)

  • Consider the set \(T = \{x \in \QQ : 2x \lt 1\}\text{.}\) Then \(T\) has no lower bound in \(\QQ\text{,}\) but it does have an upper bound and in fact a supremum: \(1/2 = \sup T\text{.}\)

  • Consider the set \(T = \{x \in \QQ : (x \geq 0) \wedge(x^2 \gt 2)\}\text{.}\) Then certainly \(1\) is a lower bound for \(T\text{,}\) but \(T\) has no infimum in \(\QQ\text{.}\)

Here's a fun observation about suprema and infima. Let \(S\) be a totally ordered set, and let \(T \subset S\) be a subset. Let \(T'\) be the set of upper bounds of \(T\text{:}\)

\begin{equation*} T' = \{x \in S : (\forall t\in T)(t\leq x)\}\text{.} \end{equation*}

Then the infimum of \(T'\) exists in \(S\) if and only if the supremum of \(T\) exists in \(S\text{,}\) in which case they are equal: \(\inf T' = \sup T\text{.}\) Dually, if \(T''\) is the set of lower bounds of \(T\text{,}\) then the supremum of \(T''\) exists if and only if the infimum of \(T\) exists, in which case they are equal. Convince yourself of this!