Infimum of Subset

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 $$\inf \left({S}\right) \preceq \inf \left({T}\right)$$.

Proof
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.