Upper Bound is Upper 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 $U$ be an upper bound for $A$.

Then $U$ is an upper bound for $B$.

Proof

Assume that:

$U$ is upper bound for $A$.

By definition of upper bound:

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

By definition of subset:

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

Hence:

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

Thus by definition

$U$ is sn upper bound for $B$.

$\blacksquare$

Sources

• Mizar article YELLOW_0:9