Infimum is not necessarily Smallest Element

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

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

Then the infimum of $T$ in $S$ is not necessarily the smallest element of $T$.

Proof
Consider the subset $T$ of the set of real numbers $\R$:
 * $T := \set {x \in \R: 1 < x \le 2}$

The number $1$ cannot be the smallest element of $T$ as $1 \notin T$.

However, $1$ is the infimum of $T$ in $S$.

Indeed, by definition:
 * $\forall x \in T: x > 1$

but let $x > 1$.

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

We have that $1 < y < x$ by Mediant is Between.

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

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

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

Hence the result.

Also see

 * Smallest Element is Infimum