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:
 * $\phi^{-1} \left({\phi \left({p}\right)^{\succ'} }\right) = p^\succ$

where $\cdot^\succ$ and $\cdot^{\succ'}$ represent strict upper closure with respect to $\preceq$ and $\preceq'$, respectively.

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

By the definition of inverse image:


 * $\phi \left({x}\right) \in \phi \left({p}\right)^{\succ'}$

By the definition of strict upper closure:


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

Since $\phi$ is an order embedding:


 * $p \prec x$

Thus by the definition of strict upper closure:


 * $x \in p^\succ$

and so:
 * $\phi^{-1} \left({\phi \left({p}\right)^{\succ'} }\right) \subseteq p^\succ$

Let $x \in p^\succ$.

By the definition of strict upper closure:


 * $p \prec x$

Since $\phi$ is an order embedding:


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

Thus by the definition of strict upper closure:


 * $\phi \left({x}\right) \in \phi \left({p}\right)^{\succ'}$

Thus by the definition of inverse image:


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

and so:
 * $p^\succ \subseteq \phi^{-1} \left({\phi \left({p}\right)^{\succ'} }\right)$

Thus by definition of set equality:
 * $\phi^{-1} \left({\phi \left({p}\right)^{\succ'} }\right) = p^\succ$