# Upper Bound for Subset

## Theorem

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

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

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

Then $U$ 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.

$\blacksquare$