“Order is one of the needs of life, which, when it is satisfied, produces a real happiness.”
―Maria Montessori
Definition1.9.Order Relation.
A total order (or linear order) is a relation \(\leq\) on a set \(S\) that is:
Reflexive: \(x\leq x\,\forall\, x\in S\)
Transitive: \(x\leq y, y\leq z\implies x\leq z \, \forall\, x,y,z\in S\)
Antisymetric: \(x\leq y, y\leq x\implies x=y\,\forall\, x,y\in S\)
Total: \(x\leq y\) or \(y\leq x\,\forall\,x,y\in S\)
A totally ordered set is a set \(S\) in which an order is defined.
Remark1.10.
The statement ''\(x < y\)'' may be read as ''\(x\) is less than \(y\)'' or ''\(x\) is smaller than \(y\)'' or ''\(x\) precedes \(y\)''.
Example1.11.
The usual orders for \(\mathbb{Z},\mathbb{Q}\text{.}\) 1
\(S=\mathbb{Z}^2=\{(a,b)\,|\, a,b\in\mathbb{Z}\}\) having lexiographic order: \((a,b)\leq (c,d)\) iff \(a<c\) or \(a=c,b\leq d\text{.}\)
Definition1.12.Bounds.
Let \(S\) be a totally ordered set, ie \(S\) is a set with a total order \(\leq\)
A subset \(E\subseteq S\) is bounded above if \(\exists\, \beta\in S\) such that \(\forall\, x\in E, x\leq \beta\text{.}\) Then \(\beta\) is called an upper bound for \(E\text{.}\)
A subset \(E\subseteq S\) is bounded below if \(\exists\, \alpha\in S\) such that \(\forall\, x\in E, \alpha\leq x\text{.}\) Then \(\alpha\) is called a lower bound for \(E\text{.}\)
A subset \(E\subseteq S\) is bounded if it is bounded from above and below.
Remark1.13.
Note \(\emptyset\) is bounded and every element of \(S\) is an upper and lower bound for \(\emptyset\) (assuming \(S\neq\emptyset\))
Definition1.14.Supremum, Infimum.
Suppose \(S\) is an ordered set, \(E\sse S\text{,}\) and \(E\) is bounded above. Suppose there exists an \(\alpha \in S\) with the following properties:
\(\a\) is an upper bound of \(E\text{.}\)
If \(y < \alpha\) then \(y\) is not an upper bound of \(E\text{.}\)
Then \(\alpha\) is called the least upper bound of \(E\) [that there is at most one such \(\alpha\) is clear from (ii)] or the supremum of \(E\text{,}\) and we write \(\a= \sup E\text{.}\) The greatest lower bound, or infimum, of a set \(E\) which is bounded below is defined in the same manner: The statement \(\a= \inf E\) means that \(\alpha\) is a lower bound of \(E\) and that no \(\b\) with \(\b > \a\) is a lower bound of \(E\text{.}\)
Theorem1.15.
When sup and inf exist, they are unique.
Example1.16.
\(S=\mathbb{Q}\) set \(E=\{x\in\mathbb{Q}\,|\, x^2<2\}\text{.}\) Has neither sup nor inf (b/c working in \(\mathbb{Q}\text{,}\) not \(\mathbb{R}\)).
\(S=\mathbb{Q}\) set \(E=\{x\in\mathbb{Q}\,|\, x>0\}\text{.}\) Then \(E\) has no sup, since not bounded above. \(\inf(E)=0\text{.}\)
Definition1.17.Least Upper Bound Property.
An ordered set \(S\) is said to have the least-upper-bound property if the following is true: If \(E\sse S\text{,}\)\(E\) is not empty, and \(E\) is bounded above, then \(\sup E\) exists in \(S\text{.}\)
Theorem1.18.Greatest Lower Bound Property.
Suppose \(S\) is an ordered set with the least-upper-bound property, \(B\sse S\text{,}\)\(B\) is not empty, and \(B\) is bounded below. Let \(L\) be the set of all lower bounds of \(B\text{.}\) Then \(\a= \sup L\) exists in \(S\text{,}\) and \(\a: = \inf B\text{.}\) In particular, \(\inf B\) exists in \(\S\text{.}\)
Since \(B\) is bounded below, \(L\) is not empty. Since \(L\) consists of exactly those \(y \in S\) which satisfy the inequality \(y \leq x\) for every \(x \in B\text{,}\) we see that every \(x \in B\) is an upper bound of \(L\text{.}\) Thus \(L\) is bounded above. Our hypothesis about \(S\) implies the refore that \(L\) has a supremum in \(S\text{;}\) call it \(\a\text{.}\)
If \(\g < \alpha\) then (see Definition 1.8) \(\g\) is not an upper bound of \(L\text{,}\) hence \(\g \not\in B\text{.}\) It follows that \(\a\leq x\) for every \(x \in B\text{.}\) Thus \(\a \in L\text{.}\)
If \(\a < \b\) then \(\a \not\in L\text{,}\) since \(\alpha\) is an upper bound of \(L\text{.}\)
We have shown that \(\alpha \in L\) but \(\b\not\in L\) if \(\b > \a\text{.}\) In other words, \(\a\) is a lower bound of \(B\text{,}\) but \(\b\) is not if \(\b > \a\text{.}\) This means that \(\a= \inf B\text{.}\)
SubsectionOrdered Fields
“There is a field beyond all notions of right and wrong. Come, meet me there.”
―Rumi
Definition1.19.Ordered Field.
An ordered field is a field \(F\) which is also an ordered set, such that
\(x + y < x + z\) if \(x, y, z \in F\) and \(y < z\text{,}\)
\(xy > 0\) if \(x \in F\text{,}\)\(y \in F\text{,}\)\(x > 0\text{,}\) and \(y > 0\text{.}\)
Definition1.20.Positive, Negative.
If \(x > 0\text{,}\) we call \(x\) positive; if \(x < 0\text{,}\)\(x\) is negative.
Theorem1.21.Properties of Ordered Fields.
The following statements are true in every ordered field.
If \(x > 0\) then \(- x < 0\text{,}\) and vice versa.
If \(x > 0\) and \(y < z\) then \(xy < xz\text{.}\)
If \(x < 0\) and \(y < z\) then \(xy > xz\text{.}\)
If \(x > 0\) then \(x^2 > 0\text{.}\) In particular, \(1 > 0\text{.}\)
If \(O < x < y\) then \(O < \frac 1/y < \frac 1/x\text{.}\)
Since \(y\leq z\text{,}\) have \(z-y\geq 0\text{.}\) Since \(x\geq 0\text{,}\) have \(x(z-y)\geq 0\implies xz-xy\geq 0\) so \(xy\leq xz\text{.}\)
If \(x\leq 0\) then \(-x\geq 0\) by (i) so by (ii) \(-xy\leq -xz\) add \(xy+xz\) to both sides, get \(xz\leq xy\text{.}\)
If \(x\geq 0\) then \(x^2=x\cdot x\geq 0\text{.}\) If \(x<0\) then by (i) \(-x>0\text{.}\) Then \(x^2=(-x)(-x)\geq 0\text{.}\)
If \(\frac{1}{y}<0\text{,}\) then \(-\frac{1}{y}>0\) as \(y>0\text{.}\)\(-1(-\frac{1}{y})y>0\) then \(1<0\) by (i), but this contradicts \(iv\) so \(\frac{1}{y}\geq 0\) since \(\frac{1}{y}\neq 0\) have \(\frac{1}{y}>0\text{.}\) By same argument \(\frac{1}{x}>0\text{.}\) Now \(\frac{1}{x}\cdot\frac{1}{y}>0\) and \(x\leq y\text{,}\) so by (ii) \(\frac{1}{x}\frac{1}{y}x\leq \frac{1}{x}\frac{1}{y}y\implies \frac{1}{x}\leq \frac{1}{y}\text{.}\)
or \(\mathbb{R}\text{,}\) although we haven't defined that yet
