Ordered Set with Order Type of Natural Numbers plus Dual has Minimum Element
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Theorem
Let $\struct {S, \preccurlyeq}$ be an ordered structure such that:
- $\map \ot {S, \preccurlyeq} = \omega + \omega^*$
where:
- $\ot$ denotes order type
- $\omega$ denotes the order type of the natural numbers $\N$
- $\omega^*$ denotes the dual of $\omega$
- $+$ denotes addition of order types.
Then $\struct {S, \preccurlyeq}$ has a smallest element.
Proof
By definition of order type addition:
- $\struct {S, \preccurlyeq}$ is isomorphic to $\struct {\N, \le} \oplus \struct {\N, \ge}$
where:
- $\cong$ denotes order isomorphism
- $\oplus$ denotes order sum.
By the Well-Ordering Principle, $\struct {\N, \le}$ has a smallest element.
By definition of order sum, every element of $\struct {\N, \le}$ precedes every element of $\struct {\N, \ge}$.
Hence the smallest element of $\struct {\N, \le}$ is also the smallest element of $\struct {\N, \le} \oplus \struct {\N, \ge}$.
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 $28 \ \text {(b)}$