4 Order
Highlights of this Chapter: We define the notion of inequality in terms of the notion of positivity which we axiomatize, leading to the definition of an ordered field. We prove this new axiom is required as not all fields can be ordered (by looking at the complex numbers), and then we investigate several important properties and definitions related to order that are essential to real analysis:
- We define absolute value, and give several characterizations
- We prove the triangle inequality
- We define square roots, and \(n^{th}\) roots
Field theory (the study of mathematical objects satisfying the field axioms) is a broad subject in mathematics, underpinning large swaths of abstract algebras as well as analysis. The first step in deciding how to order numbers is to axiomatize what it means for a number to be positive.
Definition 4.1 (Positive Elements) A subset \(P\subset \FF\) is called the positive elements if
- (Trichotomy) For every \(a\in \FF\) exactly one of the following is true: \(a=0\), \(a\in P\) \(-a\in P\).
- (Closure) If \(a,b\in P\) then \(a+b\in P\) and \(ab\in P\).
Definition 4.2 (Ordered Field) A n ordered field is a field \(\FF\) together with a subset \(P\subset \FF\) of positive elements.
Definition 4.3 (Inequalities) If \(F\) is an ordered field and \(a,b\in \FF\) then we write \(a<b\) as a shorthand for the statement that \(b-a\in P\), and we write \(a\leq b\) if either \(a<b\) or \(a=0\).
Analogously, we write \(a>b\) if \(a-b\in P\) and \(a\geq b\) if either \(a>b\) or \(a=b\).
4.0.1 Properties of Ordered Fields
Being an ordered field requires more structure above and beyond just being a field. Not every field can be ordered! In this short section we explore some properties of ordered fields in general.
Proposition 4.1 (1 is a Positive Number) If \((F,P)\) is any ordered field, then \(1\in P\).
Proof. Since \(1\neq 0\) we know that either \(1\in P\) or \(-1\in P\). So, to show \(1\in P\) its enough to see \(-1\in P\) leads to contradiction.
If \(-1\in P\) then by closure, \((-1)(-1)=1\in P\): so now we have both \(1\) and \(-1\) in \(P\), contradicting trichotomy.
Exercise 4.1 Let \(F\) be an ordered field and \(x\neq 0\) an element. Then \(x^2>0\).
Proposition 4.2 (\(\mathbb{C}\) is not ordered) The complex numbers cannot be made into an ordered field: there is no subset \(P\subset\mathbb{C}\) such that \(P\) is a positive cone for \(\mathbb{C}\).
Proof. The complex numbers contain an element \(i\) with the property that \(i^2=-1\). If they were ordered, since \(i\neq 0\) we know either \(i\in P\) or \(-i\in P\), but both of these lead to contradiction.
If \(i\in P\) then \(i^2=-1\in P\) contradicting the previous theorem that \(1\in P\) always. And \(-i\in P\) leads to the same problem: \((-i)^2=(-i)(-i)=--(i^2)=i^2=-1\), so \(-1\in P\) again.
This may seem like a strange example to start with, as the course is about real analysis. But its actually quite important: every time we introduce a new concept to the foundations of our theory we should ask ourselves, is this an axiom, or a theorem? We don’t want to add as axioms things that we can already prove from the existing axioms, as that is redundant! So before adding a new axiom, we should convince ourself that its necessary: that it is impossible to prove the existence of this new structure given the previous. And that’s what this example does. By exhibiting something that satisfies all the field axioms but cannot be ordered, we see that it is logically impossible to prove the existence of an order from the field axioms alone, and thus we must take ?def-order as a new axiom.
Exercise 4.2 The rational numbers admit a unique ordering: there is only one set \(P\subset\QQ\) that satisfies the axioms of a positive cone for \(\QQ\). Can you prove this? Hint: once we know that 1 is positive, show this determines that a fraction \(p/q\) is positive if and only if \(p\) and \(q\) have the same sign.
Some fields admit more than one possible ordering, and so can be turned into an ordered field in more than one way! An example of this is the field \(\QQ(\tau)\) with \(\tau^2=2\) admits two possible orderings, one where \(\tau\in P\) and the other with \(-\tau\in P\).
4.1 Definitions Requiring an Order
Definition 4.4 (Intervals) Let \(F\) be an ordered field. We write \([a,b]\) for the set \(\{x\mid a\leq x\leq b\}\), and call this set a closed interval in \(F\). Similarly we write \((a,b)\) for the set \(\{x\mid a<x<b\}\), which we call an open interval. Mixed intervals are also possible, such as \([a,b)=\{x\mid a\leq x <b\}\).
An unbounded interval, or a ray is a set of the form \(\{x\mid x>a\}\) or \(\{x\mid x\geq a\}\). We call the first an open ray and the latter a closed ray, and often denote them \((a,\infty)\) or \([a\infty)\) as a shorthand. Similarly with \((-\infty,a)\) and \((-\infty,a]\).
Definition 4.5 (Absolute Value) Let \(\FF\) be an ordered field. Then the absolute value is a function \(|\cdot|\colon\FF\to\FF\) defined by
\[|x|=\begin{cases} x & x\geq 0\\ -x & x<0 \end{cases}\]
Definition 4.6 (The \(\sqrt{\cdot}\) symbol) Let \(\FF\) be an ordered field, and \(x\in \FF\). If there exists a \(y\geq 0\) in \(\FF\) such that \(y^2=x\), we call \(y\) the square root of \(x\) and denote \(\sqrt{x}\).
We can generalize this by defining \(\sqrt[q]{x}\) to be the number \(y\) with \(y^p=x\), and similarly to the above, prove that if \(a<b\) are positive field elements, then \(\sqrt[q]{a}<\sqrt[q]{b}\).
Exercise 4.3 (No Square Roots of Negatives) Let \(F\) be any ordered field, and let \(x<0\). Prove that \(x\) does not have a square root in \(F\).
Definition 4.7 (Rational Powers) Let \(a\in \FF\) and \(p/q\in \QQ\).
Then if the element \(a^p\in \FF\) has a \(q^{th}\) root, we define the fractional power \(a^{p/q}\) as \[a^{p/q}=\sqrt[q]{a^p}\]
4.2 Working with Inequalities
All the standard properties of inequalities from arithmetic hold in an ordered field, and so you will be able to use them without comment throughout the course. However, its good to derive a few of these for yourselves from the definitions at first, to see how it goes.
Example 4.1 (Inequality is antisymmetric) By trichotomy we see that for every \(x\neq y\) we have either \(x<y\) or \(y<x\) (as, \(x-y\neq 0\) implies either \(x-y\in P\), so \(x-y>0\) and \(x>y\) or the reverse).
Proposition 4.3 (Inequality is transitive) Let \(F\) be an ordered field and \(a,b,c\) in \(F\). If \(a<b\) and \(b<c\), then \(a<c\).
Proof. If \(a<b\) then \(b-a\in P\). Similarly, \(b<c\) implies \(c-b\in P\). Closure then tells us their sum, \((c-b)+(b-a)\in P\), and so after simplifying, \[c+(-b+b)-a=c+0-a=c-a\in P\] This is the definition of \(c>a\).
Exercise 4.4 (Adding to an Inequality) Let \(F\) be an ordered field and \(a,b,c\in F\) with \(a<b\). Then \[a+c<b+c\]
Proposition 4.4 (Multiplying an Inequality) Let \(F\) be an ordered field and \(a,b,c\in F\) with \(a<b\). Then if \(c>0\), it follows that \(ca<cb\), and if \(c<0\) we have instead \(ca>cb\).
Proof. First treat the case \(c>0\). Since \(a<b\) we know \(b-a\in P\), and \(c\in P\) so \(c(b-a)\in P\) by the closure axiom. Distributing gives \(cb-ca\in P\) which is the definition of \(cb>ca\).
Now, if \(c<0\), we know \(c\not \in P\), so \(-c\in P\). Closure then gives \((-c)(b-a)\in P\), and simplifying yields \(-cb+ca\in P\) or \(ca-cb\in P\), the definition of \(ca>cb\).
4.2.1 Powers and Roots
Some basic inequalities for powers and roots that will prove useful: like other basic properties of inequalities, you do not need to prove or cite these when you use them in this course, but it is good to have a reference seeing why they are true from our axioms.
Example 4.2 (\(x\mapsto x^2\) is increasing) If \(F\) is an ordered field and \(a,b\in F\) are elements with \(0<a<b\) then \(a^2<b^2\).
To prove this, we use both Proposition 4.3 and Proposition 4.4. Since \(a<b\) and \(a>0\) we see \(a^2<ab\). But since \(a<b\) and \(b>0\), we see \(ab<b^2\). Putting these together yields \(a^2<ab<b^2\), so \(a^2<b^2\).
Its necessary to assume \(a,b\) are positive in the theorem above: for example \(-3<1\) but \((-3)^2=9\) is not less than \(1^2=1\). In fact this proof works in reverse as well (check this!) to provide the following useful fact:
Proposition 4.5 If \(a,b\in\FF\) are positive elements of an ordered field, then \[a<b\,\iff a^2<b^2\]
This generalizes to arbitrary powers:
Exercise 4.5 (\(x\mapsto x^n\) is increasing) Prove that if \(F\) is an ordered field containing positive elements \(a,b\), then for all \(n\in\NN\), \(a<b\) if and only if \(a^n<b^n\).
In fact, when \(n\) is odd, you may wish to prove that you can remove the assumption that \(a,b>0\).
Here’s a quick fact about inequalities that will prove useful to us later on in the course:
Exercise 4.6 (Bernoulli’s inequality) Let \(\FF\) be an ordered field and \(x>0\) be a positive element. Prove by induction that for all natural numbers \(\NN\)
\[(1+x)^n\geq 1+nx\]
Exercise 4.7 (\(\sqrt{\cdot}\) is increasing) Prove that if \(0<x<y\) in an ordered field \(F\), and \(F\) contains the square roots \(\sqrt{x},\sqrt{y}\), then \(\sqrt{x}<\sqrt{y}\).
Proposition 4.6 If \(r\in\QQ\), \(r>0\) is a positive rational number and \(x,y\in F\) are positive field elements \[x<y\implies x^r<y^r\]
Proof. Use that \(x^r=x^{p/q}=(\sqrt[q]{x^p})\) to break this into two problems: first \(x<y\) implies \(x^p<y^p\). Now, if \(u=x^p\) and \(v=y^p\) we have \(u<v\implies \sqrt[q]{u}<\sqrt[q]{v}\), completing the proof.
4.3 Working with Absolute Values
Proposition 4.7 (Absolute Values and Maxima) For all \(x\) in an ordered field, \[|x|=\max\{x,-x\}\]
Corollary 4.1 If \(x,a\) are in an ordered field, the conditions \(-x<a\) and \(x<a\) are equivalent to \[|x|<a\]
Proof. If \(-x<a\) and \(x<a\) then \(\max\{x,-x\}<a\), so by Proposition 4.7, \(|x|<a\). Conversely, if \(|x|<a\) then \(\max\{x,-x\}<a\) so both \(x<a\) and \(-x<a\).
Corollary 4.2 (Defining Feature of the Absolute Value) Let \(F\) be an ordered field: then \(|x|<a\) if and only if \(-a<x<a\).
Proof. By the above \(|x|<a\) means \(x<a\) and \(-x<a\). Multiplying the second inequality by \(-1\) yields \(x>-a\), and stringing them together results in \(-a<x<a\).
Finally, we can get a formula for the absolute value in terms of squaring and roots.
Example 4.3 For all \(x\) in an ordered field \(|x|=\sqrt{x^2}\).
Example 4.4 (Multiplication and the Absolute Value) \[|xy|=|x||y|\] \[\left|\frac{x}{y}\right|=\frac{|x|}{|y|}\]
The interaction of the absolute value with addition is more subtle, but crucial. One of the most important inequalities in all of analysis is the triangle inequality of the absolute value:
Proposition 4.8 (The Triangle Inequality) For any \(x,y\) in an ordered field \[|x+y|\leq |x|+|y|\]
Proof. It suffices to prove that we have both \[x+y\leq |x|+|y| \hspace{1cm} -(x+y)\leq |x|+|y|\]
For the first, note that as \(x\leq |x|\) and \(y\leq |y|\), \[x+y\leq |x|+y\leq |x|+|y|\] Similar reasoning succeeds for the second as \(-x\leq |x|\) and \(-y\leq |y|\): \[-x-y\leq |x|+(-y)\leq |x|+|y|\]
Exercise 4.8 Let \(a_1+a_2+\cdots + a_n\) be any finite sum. Prove that \[\left|\sum_{i=1}^n a_i\right|\leq \sum_{i=1}^n\left|a_i\right|\]
The reverse triangle inequality is another very useful property of absolute values, logically equivalent to the usual triangle inequality, but giving a lower bound for \(|a-b|\) instead of an upper bound for \(|a+b|\).
Exercise 4.9 (Reverse Triangle Inequality) Prove that for all \(a,b\) in an ordered field \(F\) \[||a|-|b||\leq |a-b|\]
Finally, two corollaries of the triangle inequality and its reverse, by replacing \(y\) with \(-y\).
Corollary 4.3 (Corollaries of the Triangle Inequality) For all \(x,y\) in an ordered field,
\[|x-y|\leq |x|+|y|\]
\[|x+y|\geq \left||x|-|y|\right|\]
4.4 \(\bigstar\) Topology
A final familiar property that arises from ordering a field is the notion of open sets and closed sets. This in turn is the foundations of the subject of topology or the abstract study of shape, which becomes quite important in advanced applications of analysis.
We will not require any deep theory in this course, and stop pause briefly to give a definition of openness and closedness.
Definition 4.8 (Open Set) A set of the form \((a,b)=\{x\mid a<x<b\}\) is called an open interval. A set \(U\subset\FF\) is called open, if for every point \(u\in U\) there is some open interval \(I\) containing \(u\) which is fully contained in \(U\): \[u\in I\subset U\]
One notable property of this definition: the empty set \(\varnothing =\{\}\) is open, as this condition is vacuously true: there are no points of \(\varnothing\) so this condition doesn’t pose any restriction!
Exercise 4.10 Explain why the set \(U=\{x\mid x>0 \textrm{ and } x\neq 2\}\) is an open set.
Exercise 4.11 Let \(\{U_n\}\) be any collection of open sets. Prove that the union \(\bigcup_n U_n\) is also open.
Hint: his collection doesn’t have to be finite, so induction won’t help us here. Can you supply a direct proof, using the definition of union and open?
Definition 4.9 (Closed Set) A set is \(K\subset \FF\) is closed, if its complement is an open set.
Exercise 4.12 Show that intervals of the form \([a,b]=\{x\mid a\leq x\leq b\}\) are closed sets. This is why we call them closed intervals in calculus courses.
This terminology is rather unfortunate when first learning the subject, as while open and closed are antonyms in english, they are not in mathematics! Being open is a special property that most sets do not have, and so being closed (which is defined relative to an open set) is also a special property. Most sets are neither open nor closed!
Example 4.5 (A set that is neither open nor closed) The set \(S=[1,2)\) is neither open nor closed. Its not open because the point \(1\in S\), but there is no open interval containing \(1\) which is fully contained in \(S\) (every open interval containing \(1\) contains numbers smaller than 1 as well).
To see its not closed, we need to show that its complement is not open. Its complement is the set \[S^c = \{x\mid x<1\}\cup\{x\mid x\geq 2\}\] Here we have the same problem at the number two: \(2\in S^c\) but there is no open interval containing \(2\) which is fully inside \(S^c\), as any such interval would contain points less than 2, and these are not in \(S^c\).
Thus, \([1,2)\) is neither open nor closed.
But perhaps even stranger, not only can sets be neither open nor closed, but they can also be both open and closed! Such sets are called clopen.
Example 4.6 (A set that is both open and closed) If \(\FF\) is the entire ordered field, \(\FF\) is both open and closed.
To see it is open, note for any \(x\) we can form the interval \((x-1,x+1)\) and this lies inside of \(\FF\). To see its closed, note that its complement is the empty set and this is vacuously open as commented above.