Preimage of Lower Section under Increasing Mapping is Lower

Theorem
Let $\left({S, \preceq}\right)$, $\left({T, \precsim}\right)$ be preordered sets.

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

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

Then $f^{-1}\left[{X}\right]$ is lower

where $f^{-1}\left[{X}\right]$ denotes the preimage of $X$ under $f$.

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

By definition of increasing mapping:
 * $f\left({y}\right) \precsim f\left({x}\right)$

By definition of preimage of set:
 * $f\left({x}\right) \in X$

By definition of lower set:
 * $f\left({y}\right) \in X$

Thus by definition of preimage of set:
 * $y \in f^{-1}\left[{X}\right]$