Supremum does not Precede Infimum

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
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