Supremum of Subset

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

Let $$\left({S; \preceq}\right)$$ admit a supremum.

Let $$T \subseteq S$$.

If $$T$$ also admits a supremum in $$S$$, then $$\sup \left({T}\right) \le \sup \left({S}\right)$$.

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

Then $$B$$ is an upper bound for $$S$$.

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

Because $$x \in S \Longrightarrow x \le B$$ (as $$B$$ is an upper bound for $$S$$) it follows that $$x \in T \Longrightarrow x \le B$$.

So $$B$$ is an upper bound for $$T$$.

Therefore $$B$$ is at least as large as the smallest upper bound (i.e. the supremum) of $$T$$ in $$S$$, hence the result.