# Infimum of Subset

## Theorem

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

Let $S \subseteq U$.

Let $T \subseteq S$.

Let $\struct {S, \preceq}$ admit an infimum in $U$.

If $T$ also admits an infimum in $U$, then $\map \inf S \preceq \map \inf T$.

## Proof

Let $B = \map \inf S$.

Then $B$ is a lower bound for $S$.

As $T \subseteq S$, it follows by the definition of a subset that $x \in T \implies x \in S$.

Because $x \in S \implies B \preceq x$ (as $B$ is a lower bound for $S$) it follows that $x \in T \implies B \preceq x$.

So $B$ is a lower bound for $T$.

Therefore $B$ precedes the infimum of $T$ in $U$.

Hence the result.

$\blacksquare$