Subset of Set is Coarser than Set

Theorem

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

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

$A \subseteq B$

Then $A$ is coarser than $B$.

Proof

Let $x \in A$.

By definition of subset:

$x \in B$

By definition of reflexivity:

$x \preceq x$

Thus

$\exists y \in B: y \preceq x$

$\blacksquare$

Sources

• Mizar article WAYBEL12:16