Lower Bound for Subset
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Theorem
Let $\left({S, \preceq}\right)$ be an ordered set.
Let $L$ be a lower bound for $S$.
Let $\left({T, \preceq}\right)$ be a subset of $\left({S, \preceq}\right)$.
Then $L$ is a lower bound for $T$.
Proof
By definition of lower bound:
- $\forall x \in S: L \preceq x$
But as $\forall y \in T: y \in S$ by definition of subset, it follows that:
- $\forall y \in T: L \preceq y$.
Hence the result, again by definition of lower bound.
$\blacksquare$