Supremum is not necessarily Greatest Element

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Theorem

Let $\struct {S, \preceq}$ be an ordered set.

Let $T$ admit a supremum in $S$.


Then the supremum of $T$ in $S$ is not necessarily the greatest element of $T$.


Proof

Proof by Counterexample:

Consider the subset $T$ of the set of real numbers $\R$:

$T := \set {x \in \R: 1 \le x < 2}$

The number $2$ cannot be the greatest element of $T$ as $2 \notin T$.

However, $2$ is the supremum of $T$ in $S$.

Indeed, by definition:

$\forall x \in T: x < 2$

So, let $x < 2$.

Then consider $y = \dfrac x 1 + \dfrac 2 1 = \dfrac {x + 2} 2$.

We have that $x < y < 2$ by Mediant is Between.

Thus $y \in T$ but $y > x$ and so $x$ cannot be the greatest element of $T$.

Neither can $y$ be the supremum of $T$ in $S$.

The conclusion is that there is no greatest element of $T$.

Hence the result.

$\blacksquare$


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