Inverse Image under Order Embedding of Strict Upper Closure of Image of Point

Theorem
Let $\left({(S, \preceq}\right)$ and $\left({T, \preceq'}\right)$ be ordered sets.

Let $\phi: S \to T$ be an order embedding of $\left({S, \preceq}\right)$ into $\left({T, \preceq'}\right)$

Let $p \in S$.

Then $\phi^{-1} \left({ {\dot\uparrow'} \phi(p) }\right) = {\dot\uparrow}p$,


 * where ${\dot\uparrow}$ and ${\dot\uparrow'}$ represent strict up-set with respect to $\preceq$ and $\preceq'$, respectively.

Proof
Let $x \in \phi^{-1} \left({ {\dot\uparrow'} \phi \left({p}\right) }\right)$.

By the definition of inverse image:


 * $\phi \left({x}\right) \in {\dot\uparrow'} \phi \left({p}\right)$

By the definition of strict up-set:


 * $\phi \left({p}\right) \prec' \phi \left({x}\right)$

Since $\phi$ is an order embedding:


 * $p \prec x$

Thus by the definition of strict up-set:


 * $x \in {\dot\uparrow}p$

Suppose instead that $x \in {\dot\uparrow} p$.

By the definition of strict up-set:


 * $p \prec x$

Since $\phi$ is an order embedding:


 * $\phi \left({p}\right) \prec' \phi \left({x}\right)$

Thus by the definition of strict up-set:


 * $\phi \left({x}\right) \in {\dot\uparrow'} \phi \left({p}\right)$

Thus by the definition of inverse image:


 * $x \in \phi^{-1} \left({ {\dot\uparrow'} \phi \left({p}\right) }\right)$

Thus by the Axiom of Extension, $\phi^{-1} \left({ {\dot\uparrow'} \phi \left({p}\right) }\right) = {\dot\uparrow}p$.