Infimum of Image of Upper Closure of Element under Increasing Mapping

Theorem
Let $\left({S, \preceq}\right)$ and $\left({T, \precsim}\right)$ be ordered set.

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

Let $x \in S$.

Then $\inf \left({f\left[{x^\succeq}\right]}\right) = f\left({x}\right)$

Proof
By Infimum of Upper Closure of Element:
 * $\inf x^\succeq = x$

By definition of infimum:
 * $x$ is lower bound for $x^\succeq$

Thus by Increasing Mapping Preserves Lower Bounds:
 * $f\left({x}\right)$ is lower bound for $f\left[{x^\succeq}\right]$

By definition of reflexivity:
 * $x \preceq x$

By definition of upper closure of element:
 * $x \in x^\succeq$

By definition of image of set:
 * $f\left({x}\right) \in f\left[{x^\succeq}\right]$

Thus by definition:
 * $\forall y \in T: y$ is lower bound for $f\left[{x^\succeq}\right] \implies y \precsim f\left({x}\right)$

Thus by definition of infimum:
 * $\inf \left({f\left[{x^\succeq}\right]}\right) = f\left({x}\right)$