Ordering/Examples/Example Ordering on Integers
Example of Ordering
Let $\preccurlyeq$ denote the relation on the set of integers $\Z$ defined as:
- $a \preccurlyeq b$ if and only if $0 \le a \le b \text { or } b \le a < 0 \text { or } a < 0 \le b$
where $\le$ is the usual ordering on $\Z$.
Then $\preccurlyeq$ is an ordering on $\Z$.
Proof
Reflexivity
Let $0 \le a$.
Then $0 \le a \le a$ and so $a \preccurlyeq a$.
Let $a < 0$.
Then $a \le a < 0$ and so $a \preccurlyeq a$.
Thus in both cases $a \preccurlyeq a$.
So $\preccurlyeq$ has been shown to be reflexive.
$\Box$
Transitivity
Let $a, b, c \in \R$ such that:
- $a \preccurlyeq b \text { and } b \preccurlyeq c$
Let $0 \le a \le b$.
Then:
- $0 \le b \le c$
and so:
- $0 \le a \le c$
That is: $a \preccurlyeq c$
Let $b \le a < 0$.
Then there are two cases for $c$:
- $(1): \quad c \le b < 0$
Then we have:
- $c \le a < 0$
That is: $a \preccurlyeq c$
- $(2): \quad b < 0 \le c$
Then we have:
- $a < 0 \le c$
That is: $a \preccurlyeq c$
In both cases:
$a \preccurlyeq c$
Let $a < 0 \le b$.
Then:
- $0 \le b \le c$
and so:
- $0 < a \le c$
That is: $a \preccurlyeq c$
In all cases:
$a \preccurlyeq c$
So $\preccurlyeq$ has been shown to be transitive.
$\Box$
Antisymmetry
Let $a, b \in \R$ such that:
- $a \preccurlyeq b \text { and } b \preccurlyeq a$
Let $0 \le a \le b$.
Then as $b \ge 0$:
- $0 \le b \le a$
Thus:
- $a \le b \text { and } b \le a$
and as $\le$ is the usual ordering, and so antisymmetric:
- $a = b$
Let $b \le a < 0$.
Then as $b < 0$:
- $a \le b < 0$
Thus:
- $a \le b \text { and } b \le a$
and as $\le$ is the usual ordering, and so antisymmetric:
- $a = b$
Let $a < 0 \le b$.
But then because $b \le a$:
- $0 \le b \le a$
which contradicts $a < 0$.
Hence if both $a \le b$ and $b \le a$, it cannot happen that $0 \le b \le a$.
All cases are accounted for, and it has been shown that:
- $a \preccurlyeq b \text { and } b \preccurlyeq a \implies a = b$
So $\preccurlyeq$ has been shown to be antisymmetric.
$\Box$
$\preccurlyeq$ has been shown to be reflexive, transitive and antisymmetric.
Hence by definition it is an ordering.
It is seen that $\preccurlyeq$ looks like:
- $-1 \preccurlyeq -2 \preccurlyeq -3 \preccurlyeq \cdots \preccurlyeq 0 \preccurlyeq 1 \preccurlyeq 2 \preccurlyeq \cdots$
$\blacksquare$
Sources
- 1996: Winfried Just and Martin Weese: Discovering Modern Set Theory. I: The Basics ... (previous) ... (next): Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: $(4)$