Infimum is not necessarily Smallest Element

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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 $V$.

$\blacksquare$


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