11. Ordered Sets

Review and Refinement:

On this Page we assume all Sets are ordered.

11.1 Segments and "Ideal Cuts":

For , we will use the notation, $\QTR{Large}{\ }$ for the segment of $\QTR{Large}{a}$ in $\QTR{Large}{A}$ . (rather than $\QTR{Large}{S(a)}$ )

The text book defines a closely related notion of an Ideal, a subset , having the property that and . I will refer to an Ideal as an Ideal Cut (Think of Dedekind Cut ) because

For any subset , is an Ideal Cut if and only if $\QTR{Large}{x<y}$ for every and

Hence if and are Ideal Cuts then or .

Also, if is a Set of Ideal Cuts, then so is
For any , $\QTR{Large}{A[a]}$ is an Ideal Cut. More generally, any subset , is an Ideal Cut if for some subset ,

We also have
Let . Show
Let , show that where the right side reads "The segment of $\QTR{Large}{t}$ in $\QTR{Large}{A[a]}$ .

Proofs:

Verify for yourself:
Exercise due March 16;
If then or . If then but by definition . If ....
In general, for we have . What we need to show is that if $\QTR{Large}{t<a}$ then

.

But

and

and

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11.2 Order morphisms :

A map is said to be an order morphism if for all , . Unless there is a possibility of confusion we will use the symbol $\QTR{Large}{\leq }$ ambiguously.

Let be a one to one order morphism then for all
Let is said to an onto order morphism then for all

or
The Segment Mapping Formula - Let be an order isomorphism (one to one, onto, and order preserving) then for all ,

Proofs:

This amounts to observing that since $\QTR{Large}{f}$ is one to one.
First note that since

(It may the the case that )

Next, let , in particular let since $\QTR{Large}{f}$ is onto let such that

If $\QTR{Large}{s<a}$ then But we cannot have $\QTR{Large}{s>a}$ since then contradicting the choice of .

Hence
Follows from 1. and 2.