Smallest Element is Infimum

Theorem

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

Let $T \subseteq S$.

Let $T$ have a smallest element $m$.

Then $m$ is the infimum of $T$ in $S$.

Proof

Let $M$ be the smallest element of $T$.

Then by definition:

$\forall x \in T: m \preceq x$

By definition of infimum, it is necessary to show that:

$(1): \quad m$ is a lower bound of $T$ in $S$

$(2): \quad L \preceq m$ for all lower bounds $L$ of $T$ in $S$.

By Smallest Element is Lower Bound, $m$ is a lower bound of $T$ in $S$.

It remains to be shown that:

$L \preceq m$ for all lower bounds $L$ of $T$ in $S$.

Let $L \in S$ be a lower bound of $T$ in $S$.

By definition of lower bound:

$\forall t \in T: L \preceq t$

We have that $m \in T$.

Therefore:

$L \preceq m$

Hence the result.

$\blacksquare$

Examples

Arbitrary Example $1$

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

$S = \set {1, 2, 3}$

Then the smallest element of $S$ is $1$.

From Smallest Element is Infimum it follows that:

$\inf S = 1$

Arbitrary Example $2$

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

$T := \set {x \in \R: 1 \le x < 2}$

$T$ has a smallest element $1$.

Hence from Smallest Element is Infimum it follows that $1$ is also the infimum of $T$.

Also see

• Greatest Element is Supremum

Sources

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