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