Total order \(\equiv\) linear order

One of the equivalent definitions of a total order reads as follows:

Notation:

• \(x \leq y\) means (\(x=y\) or \(x<y\)).

• Equivalently, \(x \leq y\) is the negation of \(x>y\).

• In Rudin’s book total order \(\equiv\) linear order is abbreviated to order.

Totally ordered sets \(\equiv\) linearly ordered sets

• In Rudin’s book totally ordered set \(\equiv\) linearly ordered set is abbreviated to ordered set.

Upper and lower bounds

Maximal/minimal and greatest/least elements

\(\sup E\) \(\equiv\) supremum of \(E\)

\(\inf E\) \(\equiv\) infimum of \(E\)

Example

Proof. Indeed, note that

• \(\frac{1}{n}\le 1\) for all \(n\in\mathbb N\), and \(1=\frac{1}{n}\in E\) for \(n=1\), thus \(1=\sup E\).

• \(\frac{1}{n}> 0\) for all \(n\in\mathbb N\). Thus \(0\) is the lower bound for \(E\), but \(0 \not\in E\).

• By the the Archimedean property for the rational numbers we know that for every positive rational number \(x>0\) there exists \(n\in \mathbb N\) such that \(0<\frac{1}{n}<x\). Thus \(\inf E=0\). $$\tag*{$\blacksquare$}$$

Solution. We have \((-1)^n=1\) for even \(n\) and \((-1)^n=-1\) for odd \(n\).

Hence \[E=\{-1,1\}.\]

Consequently \[\begin{aligned} \sup E&=\max E=1,\\ \inf E&=\min E=-1. \end{aligned}\] $$\tag*{$\blacksquare$}$$

Solution. Note that

• \(\frac{1}{n^2+1} \leq \frac{1}{1}\) for all \(n \in \mathbb{N}_0\), and \(1=\frac{1}{n^2+1}\in E\) for \(n=0\), hence \(1\) is the greatest element of \(E\) and we have \(\sup E=1\).

• \(\frac{1}{n^2+1}>0\) for all \(n\in \mathbb{N}_0\). Thus \(0\) is the lower bound for \(E\), but \(0 \not\in E\).

• By the the Archimedean property for the rational numbers we know that for every positive rational number \(x>0\) there exists \(n\in \mathbb N\) such that \(0<\frac{1}{n}<x\). Also \(n<n^2+1\) which implies \(0<\frac{1}{n^2+1}<\frac{1}{n}<x\). Hence \(\inf E=0\not\in E\).

Least–upper–bound property

Theorem

Proof. Let \[L=\{y \in S\colon y \leq x \text{ for all }x \in B\}.\]

We see that \(L \neq \varnothing\), since \(B\) is bounded below. Every \(x \in B\) is an upper bound of \(L\). Thus \(L\) is bounded above and consequently the least–upper–bound property implies that \(\alpha=\sup L\) exists in \(S\).

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We show that \(\alpha \in L\). It suffices to prove that \(\alpha \leq x\) for all \(x \in B\). Suppose for a contradiction that there is \(\gamma\in B\) such that \(\gamma< \alpha\). By the definition of supremum \(\gamma\) is not an upper bound. Therefore, there exists \(y \in L\) such that \(\gamma<y\leq \alpha\), so \(y \leq x\) for every \(x \in B\), and hence \(\gamma<x\) for all \(x \in B\). In particular, we obtain \(\gamma<\gamma\) since \(\gamma\in B\), which is impossible!

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Now we show that \(\alpha=\inf B\). We have shown that \(\alpha \in L\), which means that \(\alpha\) is a lower bound of \(B\), since \(\alpha\le x\) for all \(x\in B\). If \(\alpha < \beta\), then \(\beta \not\in L\). If not, we would have \(\alpha < \beta\le \alpha=\sup L\). Since \(\beta\not\in L\) then there exists \(x\in B\) such that \(\beta> x\ge \alpha\). This proves that \(\alpha=\inf B\). $$\tag*{$\blacksquare$}$$

Field 1/2

Field 2/2

Field properties - addition

Proof of (A). \[\begin{aligned} y&\overbrace{=}^{\color{red}(A4)} 0_{\mathbb{F}}+y \overbrace{=}^{\color{red}(A5)} (-x+x)+y \overbrace{=}^{\color{red}(A3)} -x+(x+y)\\& \ =-x+(x+z) \overbrace{=}^{\color{red}(A3)} (-x+x)+z \overbrace{=}^{\color{red}(A5)} 0_{\mathbb{F}}+z \overbrace{=}^{\color{red}(A4)} z. \end{aligned}\]

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To prove (B), we take \(z=0_{\mathbb{F}}\) in (A).

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To prove (C) we take \(z=-x\) in (A).

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Since \(x+(-x)=0_{\mathbb{F}}\), so by (C) with \(-x\) in place of \(x\) we get \[(-(-x))=x.\] $$\tag*{$\blacksquare$}$$

Field properties - multiplication

Exercise.

Further field properties

• For the proof of (A), we use (D1): \[0_{\mathbb{F}}x+0_{\mathbb{F}}x\overbrace{=}^{{\color{brown}(D1)}}(0_{\mathbb{F}}+0_{\mathbb{F}})x=0_{\mathbb{F}}x.\] Thus we must have \(0_{\mathbb{F}}x=0_{\mathbb{F}}\).

• To prove (B) assume \(x,y \neq 0_{\mathbb{F}}\), but \(xy=0_{\mathbb{F}}\). Then \[1_{\mathbb{F}}={x}^{-1}{y}^{-1}xy={x}^{-1}{y}^{-1}0_{\mathbb{F}}=0_{\mathbb{F}},\] but \(0_{\mathbb{F}} \neq 1_{\mathbb{F}}\).

• To prove (C) we write \[(-x)y+xy\overbrace{=}^{{\color{brown}(D1)}}(-x+x)y=0_{\mathbb{F}}y=0_{\mathbb{F}},\] thus \((-x)y=-(xy)\).

• To prove \((D)\) we use \((C)\) and we write \[(-x)(-y)=-(x(-y))=-(-(xy))=xy.\] $$\tag*{$\blacksquare$}$$

Ordered fields

Properties of ordered fields

Proof of (A).

• If \(x>0_{\mathbb{F}}\), then \(0_{\mathbb{F}}=-x+x>-x+0_{\mathbb{F}}\), thus \(-x<0_{\mathbb{F}}\).

• If \(x<0_{\mathbb{F}}\), then \(0_{\mathbb{F}}=-x+x<-x+0_{\mathbb{F}}\), so that \(-x>0_{\mathbb{F}}\).

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Proof of (B).

• Since \(z>y\) we have \(z-y>y-y=0_{\mathbb{F}}\), hence \(x(z-y)>0_{\mathbb{F}}\) if \(x>0_{\mathbb{F}}\).

• Thus \[xz=x(z-y)+xy>0_{\mathbb{F}}+xy=xy.\]

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Proof of (C). By (A),(B), and \((-x)y=-(xy)=x(-y)\): \[-(x(z-y))=(-x)(z-y)>0_{\mathbb{F}}\] so that \(x(z-y)<0_{\mathbb{F}}\) hence \(xz<xy\).

Proof of (D).

• If \(x>0_{\mathbb{F}}\) we get \(x^2>0_{\mathbb{F}}\).

• If \(x<0_{\mathbb{F}}\), then \(-x>0_{\mathbb{F}}\), hence \((-x)^2>0_{\mathbb{F}}\), but \(x^2=(-x)^2\).

• We also see \(1_{\mathbb{F}}^2=1_{\mathbb{F}}\), thus \(1_{\mathbb{F}}>0_{\mathbb{F}}\).

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Proof of (E).

• If \(y>0_{\mathbb{F}}\) and \(v \leq 0_{\mathbb{F}}\), then \(yv \leq 0_{\mathbb{F}}\).

• But \({y}^{-1}\cdot y=1_{\mathbb{F}}>0_{\mathbb{F}}\), thus \({y}^{-1}>0_{\mathbb{F}}\).

• In similar way \({x}^{-1}>0_{\mathbb{F}}\).

• Multiplying the inequality \(x<y\) by \({x}^{-1}{y}^{-1}\) we have \[0_{\mathbb{F}}<{y}^{-1}<{x}^{-1}.\]

$$\tag*{$\blacksquare$}$$

A useful lemma

The absolute value in ordered fields

• If \(x\ge 0\) and \(y\ge 0\) then \(xy\ge0\) and \(|xy|=xy=|x||y|\). If \(x\ge 0\) and \(y< 0\) then \(xy\le 0\) and \(|xy|=-(xy)=x \cdot (-y)=|x||y|\). Two other cases \(x<0\) and \(y\ge0\) or \(x<0\) and \(y<0\) can be covered similarly.

• Clearly \(x\le |x|\) for all \(x\in \mathbb R\). Similarly, \(-x\le |x|\) giving \(x\ge-|x|\).

• Since \(x\le |x|\) and \(y\le |y|\), then \(x+y\le |x|+|y|\). We also have \(-(|x|+|y|)\le x+y\), since \(-|x|\le x\) and \(-|y|\le y\). Hence \[-( |x|+|y|)\le x+y\le |x|+|y| \qquad \Longleftrightarrow\qquad |x+y| \leq |x|+|y|.\]

• Note that \({\color{red}|x|=|y+x-y|\le |y|+|x-y|,}\) and similarly we obtain \({\color{blue}|y|=|x+y-x|\le |x|+|x-y|.}\) Thus \[\begin{aligned} {\color{blue}-|x-y|\le |x|-|y|} \quad \text{ and } \quad {\color{red}|x|-|y|\le |x-y|}, \end{aligned}\] which gives \(||x|-|y|| \leq |x-y|\).$$\tag*{$\blacksquare$}$$

Subfields and field homomorphisms

Field embeddings

An injective ordered field homomorphism should be thought of as an embedding. We will show that \(\mathbb Q\) can be realized as a subset of any ordered field \(\mathbb F\), in a way that respects all the ordered field structure.

• First we must check that \(\varphi\) is well defined: if \(\frac{m_1}{n_1}=\frac{m_2}{n_2}\), then \(m_1n_2=m_2n_1\). By an easy induction it follows that \[(m_1\cdot 1_{\mathbb F})\cdot (n_2\cdot 1_{\mathbb F})=(m_2\cdot 1_{\mathbb F})\cdot (n_2\cdot 1_{\mathbb F}).\] Dividing out on both sides then shows that \(\varphi\) is well-defined, since \[\varphi\bigg(\frac{m_1}{n_1}\bigg)=(m_1\cdot 1_{\mathbb F})\cdot (n_1\cdot 1_{\mathbb F})^{-1}=(m_2\cdot 1_{\mathbb F})\cdot (n_2\cdot 1_{\mathbb F})^{-1}= \varphi\bigg(\frac{m_2}{n_2}\bigg)\]

• It is routine to verify that \(\varphi\) is an ordered field homomorphism.

• Finally, to show that \(\varphi\) is injective, suppose that \(\varphi(q_1)=\varphi(q_2)\) this means, using the homomorphism property, that \(\varphi(q_1-q_2)=0_{\mathbb F}\). Let \(q_1-q_2=\frac{m}{n}\), then \(0_{\mathbb F}=\varphi(q_1-q_2)=\varphi\big(\frac{m}{n}\big)=(m\cdot 1_{\mathbb F})\cdot (n\cdot 1_{\mathbb F})^{-1}\), which implies that \(m\cdot 1_{\mathbb F}=0_{\mathbb F}\). By the previous lemma this in turn implies that \(m=0\). Thus \(q_1=q_2\) as desired and \(\varphi\) is injective. $$\tag*{$\blacksquare$}$$