Lower Bound is Lower Bound for Subset

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Theorem

Let $\left({S, \preceq}\right)$ be a preordered set.

Let $A, B$ be subsets of $S$ such that

$B \subseteq A$

Let $L$ be an element of $S$.

Let $L$ be a lower bound for $A$.


Then $L$ is a lower bound for $B$.


Proof

Let $L$ be a lower bound for $A$.

By definition of lower bound:

$\forall x \in A: L \preceq x$

By definition of subset:

$\forall x \in B: x \in A$

Hence:

$\forall x \in B: L \preceq x$

Thus by definition:

$L$ is a lower bound for $B$.

$\blacksquare$


Sources