Upper Bound for Subset
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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.
$\blacksquare$