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

Let $V$ be the subset of the real numbers $\R$ defined as:

$V := \set {x \in \R: x > 0}$

From Infimum of Subset of Real Numbers: Example 3, $V$ admits an infimum:

$\inf V = 0$

But $V$ has no smallest element, as follows.

We note that $\inf V = 0 \notin V$.

Aiming for a contradiction, suppose $x \in V$ is the smallest element of $V$.

Then $\dfrac x 2$ is also in $V$ as $\dfrac x 2 > 0$.

But $\dfrac x 2 < x$ which contradicts $x$ as being the smallest element of $V$.

Hence there can be no smallest element of $V$.

$\blacksquare$

Also see

• Smallest Element is Infimum

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 2$: Continuum Property: $\S 2.7$: Maximum and Minimum