Supremum does not Precede Infimum
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Theorem
Let $\struct {S, \preceq}$ be an ordered set.
Let $T \subseteq S$ admit both a supremum $M$ and an infimum $m$.
Then $m \preceq M$.
Proof
By definition of supremum:
- $\forall a \in T: a \preceq M$
By definition of infimum:
- $\forall a \in T: m \preceq a$
The result follows from transitivity of ordering.
$\blacksquare$
Sources
- 1947: James M. Hyslop: Infinite Series (3rd ed.) ... (previous) ... (next): Chapter $\text I$: Functions and Limits: $\S 3$: Bounds of a Function