Dedekind Completeness is Self-Dual

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Theorem

Let $\left({S, \preceq}\right)$ be an ordered set.


Then $\left({S, \preceq}\right)$ is Dedekind complete iff every non-empty subset of $S$ that is bounded below admits an infimum in $S$.


That is, an ordered set is Dedekind complete iff its dual is Dedekind complete.


Proof

Necessary Condition

Let $\left({S, \preceq}\right)$ be Dedekind complete.

Let $A \subseteq S$ be non-empty and bounded below.


Let $B \subseteq S$ be set of all lower bounds for $A$.

Then every element of $A$ is an upper bound for $B$.

Therefore, $B$ is non-empty and bounded above.

By the definition of Dedekind completeness, $B$ admits a supremum $x \in S$.


By the definition of supremum, it follows that every element of $A$ succeeds $x$.

That is, $x$ is a lower bound for $A$.

If $y \in S$ is a lower bound for $A$, then $y \in B$, and so $y \preceq x$.

Hence, $x$ is the infimum of $A$.

$\Box$


Sufficient Condition

Follows directly from the Necessary Condition and Dual of Dual Ordering.

$\blacksquare$


Sources