# Upper Bound is Upper Bound for Subset

## 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