Dual of Total Ordering is Total Ordering

Theorem

Let $\preccurlyeq$ be a total ordering.


Then its dual ordering $\succcurlyeq$ is also a total ordering.


Proof

Let $\struct {S, \preccurlyeq}$ be a totally ordered set.

From Dual Ordering is Ordering we have that $\succcurlyeq$ is an ordering.


Let $x, y \in S$.

Then $x \preccurlyeq y$ or $y \preccurlyeq x$ since $\preccurlyeq$ is total.

By definition of dual ordering, also $y \succcurlyeq x$ or $x \succcurlyeq y$.

Hence the result, by definition of total ordering.

$\blacksquare$


Sources

  • 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Theorem $14.4$
  • 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.10$
  • 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 $27 \ \text {(b)}$
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