Smallest Element is Infimum

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\left({S, \preceq}\right)$ 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

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:

$\inf S = 1$


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


Sources