Supremum is not necessarily Greatest Element/Proof
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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
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.
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 2$: Continuum Property: $\S 2.8$: Example $\text{(ii)}$