Infimum of Subset

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Theorem

Let $\left({S, \preceq}\right)$ be an ordered set.

Let $\left({S, \preceq}\right)$ admit an infimum.

Let $T \subseteq S$.

If $T$ also admits an infimum in $S$, then $\inf \left({S}\right) \preceq \inf \left({T}\right)$.

Let $B = \inf \left({S}\right)$.

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 $S$, hence the result.

$\blacksquare$