Upper Bound for Subset

Theorem
Let $$\left({S; \preceq}\right)$$ be a poset.

Let $$U$$ be an upper bound for $$S$$.

Let $$\left({T; \preceq}\right)$$ be a subset of $$\left({S; \preceq}\right)$$.

Then $$L$$ is an upper bound for $$T$$.

Proof
By definition of upper bound:
 * $$\forall x \in S: x \preceq U$$

But as $$\forall y \in T: y \in S$$ by definition of subset, it follows that:
 * $$\forall y \in T; y \preceq U$$.

Hence the result, again by definition of upper bound.

Also see

 * Lower Bound for Subset