Preimage of Lower Section under Increasing Mapping is Lower

Theorem
Let $\struct {S, \preceq}$, $\struct {T, \precsim}$ be preordered sets.

Let $f: S \to T$ be an increasing mapping.

Let $X \subseteq T$ be a lower subset of $T$.

Then $f^{-1} \sqbrk X$ is lower

where $f^{-1} \sqbrk X$ denotes the preimage of $X$ under $f$.

Proof
Let $x \in f^{-1} \sqbrk X$, $y \in S$ such that
 * $y \preceq x$

By definition of increasing mapping:
 * $\map f y \precsim \map f x$

By definition of preimage of set:
 * $\map f x \in X$

By definition of lower set:
 * $\map f y \in X$

Thus by definition of preimage of set:
 * $y \in f^{-1} \sqbrk X$