Simple Order Product/Examples/Unit Square with Open Side
Example of Simple Order Product
Consider the simple order product of the real intervals $\hointr 0 1$ and $\closedint 0 1$ under the usual ordering:
- $\struct {T, \preccurlyeq_s} := \struct {\hointr 0 1, \le} \otimes^s \struct {\closedint 0 1, \le}$
$\struct {T, \preccurlyeq_s}$ has one minimal element:
- $\tuple {0, 0}$
which is also the smallest element: of $\struct {T, \preccurlyeq_s}$.
$\struct {T, \preccurlyeq_s}$ has no greatest element and no maximal elements.
Proof
Recall the definition of the ordering in the simple order product:
- $\forall \tuple {a, b}, \tuple {c, d} \in T: \tuple {a, b} \preccurlyeq_s \tuple {c, d} \iff a \le c \text { and } b \le d$
Let $\tuple {a, b} \in T$ be a minimal element of $T$.
- $\forall \tuple {x, y} \in T: \tuple {x, y} \preccurlyeq_s \tuple {a, b} \implies \tuple {x, y} = \tuple {a, b}$
Suppose $a \ne 0$.
Then:
- $\tuple {0, b} \preccurlyeq_s \tuple {a, b}$ but $\tuple {0, b} \ne \tuple {a, b}$
Similarly, suppose $b \ne 0$.
Then:
- $\tuple {a, 0} \preccurlyeq_s \tuple {a, b}$ but $\tuple {a, 0} \ne \tuple {a, b}$
Hence if $\tuple {a, b}$ is a minimal element of $T$ it follows that:
- $\tuple {a, b} = \tuple {0, 0}$
$\Box$
Next we note that:
- $\forall \tuple {c, d} \in T: 0 \le c, 0 \le d$
and so:
- $\forall \tuple {c, d} \in T: \tuple {0, 0} \preccurlyeq_s \tuple {c, d}$
Hence by definition $\tuple {0, 0}$ is a smallest element of $T$.
By Smallest Element is Unique, $\tuple {0, 0}$ is then the smallest element of $T$.
$\Box$
Aiming for a contradiction, suppose $T$ has a maximal element $\tuple {u, v}$.
Then we have:
- $\tuple {u, v} \preccurlyeq_s \tuple {x, y} \implies \tuple {u, v} = \tuple {x, y}$
In particular, this means:
- $u \le x \implies u = x$
But we have that:
- $u \in \hointr 0 1$
and so:
- $u < 1$
So take:
- $x = \dfrac u 1 + \dfrac 1 1 = \dfrac {u + 1} 2$
Hence from Mediant is Between:
- $u < x$
and so it is not the case that $u \le x \implies u = x$.
Hence it is not the case that:
- $\tuple {u, v} \preccurlyeq_s \tuple {x, y} \implies \tuple {u, v} = \tuple {x, y}$
and so $\tuple {u, v}$ is not a maximal element of $T$.
So by Proof by Contradiction it follows that there is no maximal element of $T$.
$\Box$
Aiming for a contradiction, suppose $\tuple {u, v}$ is a greatest element of $T$.
From Greatest Element is Maximal, $\tuple {u, v}$ would have to be a maximal element of $T$.
But from above $T$ has no maximal element.
Hence by Proof by Contradiction it follows that there is no greatest element of $T$.
$\Box$
Hence the result.
$\blacksquare$
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: Exercise $31$